 # Geometrical Optics

## Concept Of Physics

### H C Verma

1   1. A concave mirror having a radius of curvature 40 cm is placed in front of an illuminated point source at a distance of $30 cm$ from it. Find the location of the image.

##### Solution : $u$ = -30 cm, R = -40 cm, $\\$ from the mirror equation.$\\$ $\frac{1}{v}$ + $\frac{1}{u}$ = $\frac{2}{R}$ $\\$ Then $\frac{1}{v}$ = $\frac{2}{R}$ - $\frac{1}{ u}$ = $\frac{2}{-40}$ - $\frac{1}{-30}$ = $\frac{1}{60}$ $\\$ or, v = -60 cm,$\\$ So, the image will be formed at a distance of 60 cm in front of the mirror.

2   2. A concave mirror forms an image of $20 cm$ high object on a screen placed $5'0 m$ away from the mirror. The height of the image is $50 cm$. Find the focal length of the mirror and the distance between the mirror and the object.

##### Solution : Given that. $\\$ ${H_1 }$= 20 cm, v = -5 m = -500 cm,${ h_2}$ = 50 cm.$\\$ Since, $\frac{-V}{u}$ = $\frac{h_2}{h_1}$ $\\$ or $\\$ $\frac{500}{u}$ = -$\frac{50}{20}$(because the image inverted).$\\$ or u = -$\frac {500\times2}{5}$ = -200 cm = -2 m. $\\$ $\frac{1}{v}$ + $\frac{1 }{u }$= $\frac{ 1}{f}$ or $\frac{ 1}{ -5 }$ + $\frac{ 1}{-2 }$ = $\frac{ 1}{f }$ $\\$ or f = $\frac{ -10}{7 }$ = -1.44 m.$\\$ So,the focal length is 144 m.

3   3. A concave mirror has a focal length of $20 cm$. Find the position or positions of an object for which the imagesize is double of the object-size.

##### Solution : For the concave miror,f = -20 cm,M = $\frac{v}{u}$ = 2.4$\\$ then v = -2u.$\\$ first case $\\$ $\frac{1 }{v }$ + $\frac{ 1}{ u}$ = -$\frac{1 }{f }$ $\\$ then u = $\frac{f }{2 }$ = 10 cm.$\\$ second case $\\$ $\frac{ -1}{2u }$ - $\frac{ 1}{ u}$ = $\frac{ -1}{f }$ $\\$ then $\frac{ 3}{ 2u}$ = $\frac{1 }{f }$$\\ then u = \frac{3f }{2 } = 30 cm. \\ the positions are 10 cm or 30 cm from the concave mirror. 4 4. A 1 cm object is placed perpendicular to the principal axis of a convex mirror of focal length 7'5 cm. Find its distance from the mirror if the image formed is 0.6 cm4 in size. ##### Solution : m = \frac{ -v} {u } = 0.6 and f = 75 cm = \frac{15 }{ 2} cm \\ From mirror equation, \frac{ 1}{ v } + \frac{ 1}{u } = \frac{1 }{f } \\ then \frac{1 }{0.6u } - \frac{ 1}{u } = \frac{1 }{f }$$\\$ u = 5 cm.

5   5. A candle flame $1'6 cm$high is imaged in a ball bearing of diameter $0'4 cm$. If the ball bearing is $20 cm$away from the flame, find the location and the height of the image.

##### Solution : Height of the object AB = 1.6 cm,$\\$ Diameter of the ball bearing = d = 0.4 cm. $\\$ then R = 0.2 cm, $\\$ Given, u = 20 cm.$\\$ We know $\frac{ 1} { u}$ + $\frac{1 } {v }$ = $\frac{ 2} {r }$ $\\$ putting the values acording to the sign conventions.$\frac{1 } {-20 }$ + $\frac{1 } { v}$ = $\frac{ 2} { 0.2}$ $\\$ $\frac{ 1} {v }$ = $\frac{v } {20 }$ + 10 = $\frac{201 } {20 }$ = v = 0.1 cm=1 mm inside the ball bearing, $\\$ magnification= m = $\frac{(AB)^1 } {AB}$ =- $\frac{ v} { u}$ = -$\frac{ 0.1} {-20 }$ = $\frac{1 } { 200} $$\\ then A^{1}b^{1 }= \frac{AB } {200 } = \frac{ 16} { 200} = 0.08 cm = 0.8 m. 6 6. A 3 cm tall object is placed at a distance of 7.5 cm from a convex mirror of focal length 6 cm. Find the location, size and nature of the image. ##### Solution : Given AB = 3 cm, u = -7.5 cm, f = 6 cm,\\ using \frac{1 } {v } + \frac{1 } { u} = \frac{1 } { f} \Rightarrow \frac{1 } { v } = \frac{1 } { f } - \frac{1 } { u}$$\\$ putting the values according to the sign conventions.$\\$ $\frac{ 1} { v }$ = $\frac{1 } { 6}$ - $\frac{1 } {-7.5 }$ = $\frac{ 3} { 10 }$ $\\$ v = $\frac{10 } { 3 } cm$ $\\$ then magnification = m =- $\frac{ v} { u}$ = $\frac{ 10} {7.5\times3 }$$\\ then \frac{ A^1B^1} { AB} = \frac{10 } { 7.5\times 3 }$$\\$ then ${A^1B^1}$ = $\frac{ 100} { 72 }$ = $\frac{ 4}{ 3}$ = 1.33 cm.$\\$ $\therefore$ image will form at a distance of $\frac{10 } {3 }$ cm.From the pole and image is 1.33 cm(virtual and erect).

7   7. A U-shaped wire is placed before a concave mirror having radius of curvature $20 cm$ as shown in figure (18-E1). Find the total length of the image.

##### Solution : R = 20 cm, f = $\frac{ R}{2 }$ = -10 cm.$\\$ For part AB, PB = 30+10 = 40 cm $\\$ So, u = -40 cm $\\$ then $\frac{1 }{v }$ = $\frac{1 }{f }$ - $\frac{1 }{u }$ = -$\frac{1 }{ 10}$ -$\big(\frac{ 1}{-40 }\big)$=$-\frac{3}{40}$ $\\$ then v = - $\frac{40 }{3 }$ = -1.33 cm.$\\$ So. $PB^{1}$ = 13.3cm$\\$ m = $\frac{ A^1B^1}{AB }$ = -$\frac{v }{u }$ = $-$$\big(\frac{ -13.3}{-40 }\big) = -\frac{1 }{3 } \\ Then A^{1}B^{1} = \frac{-10 }{3 } = -3.33 cm\\ for part CD,PC = 30, So , u= -30 cm\\ \frac{ 1}{v} = \frac{ 1}{f } - \frac{ 1}{u } = -\frac{ 1}{ 10} - \big(-\frac{1}{30}\big) = -\frac{1 }{5 }cm \\ then v = -15 cm = PC^{1}$$\\$ So, m = $\frac{C^{1}D^{1}}{ CD}$ = - $\frac{v }{ u}$ = $-$ $\big(\frac{-15}{-30}\big)$ =- $\frac{ 1}{ 2}$$\\ then C^{1}D^{1} = 5 cm.\\ B^{1}C^{1} = PC^{1} - PB^{1} = 15 - 13.3 = 17 cm.\\ So, total length A^{1}B^{1} + B^{1}C^{1} + C^{1}D^{1} = 3.3 + 1.7 + 5 = 10 cm. 8 8. A man uses a concave mirror for shaving. He keeps his face at a distance of 25 cm from the mirror and gets an image which is 1'4 times enlarged. Find the focal length of the mirror. ##### Solution : U = -25 cm.\\ m = \frac{A^{1}B^{1} }{ AB} = - \frac{ v}{u }$$\\$ then 1.4 = -$\big(\frac{v}{-25}\big)$ $\\$ then $\frac{ 14}{ 10}$ = $\frac{ v}{ 25}$ $\\$ v = $\frac{ 25\times14}{10 }$ = 35 cm.$\\$ Now , $\frac{1 }{v }$ + $\frac{ 1}{u }$ = $\frac{1 }{f }$$\\ then \frac{ 1}{ f} = \frac{ 1}{ 35} - \big(\frac{1}{-25}\big) = \frac{5 - 7 }{175 } = - \frac{ 2}{ 175}$$\\$ then f = -87.5 cm.$\\$ So focal length of the concave mirror is 87.5cm,

9   9. Find the diameter of the image of the moon formed by a spherical concave mirror of focal length $7'6 m$. The diameter of the moon is $3450 km$ and the distance between the earth and the moon is $3.8$ x $10 5 km4$.

##### Solution : u = $-3.8\times10^{5}$ km $\\$ diameter of moon = 3450 km ; f = -7.6 m $\\$ $\frac{ 1}{ v}$ + $\frac{ 1}{u }$ = $\frac{ 1}{ f}$$\\ then \frac{1 }{ v} + \big(-\frac{1}{3.8\times10^{5}}\big) = \big(-\frac{1}{7.6}\big)$$\\$ since the distance of moon from the earth is very large as compared to focal length it can taken as infinty.$\\$ image will be formed at focus which is inverted.$\\$ $\frac{1 }{ v}$ = -$\big(\frac{1}{7.6}\big)$ $\\$ then v = -7.6m$\\$ m = - $\frac{v }{u }$ = $\frac{ d_{image}}{ d_{object}}$ $\\$ then = - $\frac{- (-76)}{(-3.8\times10^{8}) }$ = $\frac{ d_{image}}{ 3450\times10^{3}}$$\\ d_{image} = \frac{ 3450\times7.6\times10^{3}}{ 3.8\times10^{8}} = 0.069 m = 6.9 cm\\ 10 10. A particle goes in a circle of radius 2'0 cm. A concave mirror of focal length 20 cm is placed with its principal axis passing through the centre of the circle and perpendicular to its plane. The distance between the pole of the mirror and the centre of the circle is 30 cm. Calculate the radius of the circle formed by the image. ##### Solution : u = -30cm, f = -20 cm\\ We know \frac{ 1}{v } + \frac{1 }{u } = \frac{1 }{f } \\ \frac{ 1}{v } + \big(-\frac{1}{30}\big) = \big(-\frac{1}{20}\big) \\ then v = -60 cm. \\ image of the circle is formed at a distance 60cm in front of the mirror.\\ m = -\frac{ v}{u } = - \frac{-60 }{ -30} = \frac{ R_{image}}{ R_{object}}$$\\$ then $\frac{ R_{image}}{2 }$$\\ then R_{image} = 4cm.\\ radius of image of the circle is 4 cm. 11 11. A concave mirror of radius R is kept on a horizontal table (figure 18-E2). Water (refractive index = p) is poured into it upto a height h. Where should an object be placed so that its image is formed on itself ? ##### Solution : Let the object placed at a height x above the surface of water.\\ The apparent position of the object with respect to mirror should be at the center of curvature so that the image is formed at the same position.\\ Since, \frac{ Real depth}{Apparent depth } = \frac{ 1}{ \mu}(with respect to mirror)\\ Now, \frac{ x}{ R - h} = \frac{1 }{\mu }$$\\$ then x = $\frac{ R - h}{\mu }$.

12   12. A point source $S$ is placed midway between two converging mirrors having equal focal length $f$ as shown in figure $(18-E3)$. Find the values of d for which only one image is formed.

##### Solution : Both the mirrors have equal focal length f.$\\$ They will produce one image under two conditions.$\\$ Case 1 : When the source is at distance '2f' from each mirror.that is the source is at center of curvature of the mirrors. The image will be produced at the same point S.$\\$ So. d = 2f + 2f = 4f.$\\$ Case 2 : When the source s ia at a distance f from each mirror.the rays from the source after reflecting from one mirror will become parallel and so these parallel rays after the reflection from the other mirror the object itself. So. only sine image is formed.$\\$ Here f + f = 2f.$\\$

13   13. A converging mirror $M1$, a point source $S$ and a diverging mirror $M2$ a re arranged as shown in figure (18-E4). The source is placed at a distance of $30 cm$ from $M1$. The focal length of each of the mirrors is $20 cm$. Consider only the images formed by a maximum of two reflections. It is found that one image is formed on the source itself. $(a)$ Find the distance between the two mirrors. $(b)$ Find the location of the image formed by the single reflection from $M$, .

##### Solution : As shown in figure ,for first reflection im $M_1$, u = -30 cm,f = -20cm$\\$ $\frac{1 }{v }$ + $\frac{1 }{30 }$ = - $\frac{1 }{20 }$$\\ then v = -60 cm.\\ So for second reflection in M_2$$\\$ u = 60 - (30 + x) = 30 - x$\\$ v = -x; f = 20cm$\\$ then $\frac{ 1}{30 - x }$ - $\frac{ 1}{ x}$ = $\frac{ 1}{20 }$ Then $x^2$ + 10x -600 = 0$\\$ Then x = $\frac{ 10\pm50}{2 }$ = $\frac{ 40}{2 }$ = 20 cm or -30 cm $\\$ Then Total distance between the two lines is 20 + 30 = 50 cm.

14   14. A light ray falling at an angle of $45°$ with the surface of a clean slab of ice of thickness $1'00 m$ is refracted into it at an angle of $30°$. Calculate the time taken by the light rays, to cross the slab. Speed of light in vacuum - $3$ x $10^{8}$m/s.

##### Solution :

We know , $\frac{sin i}{sin r}$ = $\frac{ 3\times10^{8}}{v }$ = $\frac{sin 45^0 }{ sin 30^0}$ = $\sqrt2$ $\\$ Then v = $\frac{3 \times 10^8 }{\sqrt2 }$ $\frac{ m}{ sec}$ $\\$ Distance travelled by light in the slab is. $\\$ x = $\frac{1 m }{cos 30^0 }$ = $\frac{ 2}{\sqrt3 }$ m. $\\$ So timetaken = $\frac{ 2\times \sqrt2}{ \sqrt3\times3\times\times10^8}$ = 0.54 x 10^{-8} = 5.4 x 10^{-9} sec. We know , $\frac{sin i}{sin r}$ = $\frac{ 3\times10^{8}}{v }$ = $\frac{sin 45^0 }{ sin 30^0}$ = $\sqrt2$ $\\$ Then v = $\frac{3 \times 10^8 }{\sqrt2 }$ $\frac{ m}{ sec}$ $\\$ Distance travelled by light in the slab is. $\\$ x = $\frac{1 m }{cos 30^0 }$ = $\frac{ 2}{\sqrt3 }$ m. $\\$ So timetaken = $\frac{ 2\times \sqrt2}{ \sqrt3\times3\times\times10^8}$ = $0.54$ x $10^{-8}$ = $5.4$ x $10^{-9}$ sec.

15   16. A small piece of wood is floating on the surface of a $2.5 m$ deep lake. Where does the shadow form on the bottom when the sun is just setting? Refractive index of water = $4/3$.

##### Solution : Height of the lake = 2.5 m $\\$ When the sun is just setting , $\theta$ is approximately = $90 ^0$$\\ Then \frac{sin i }{sin r } = \frac{ \mu_2}{\mu_1 }$$\Rightarrow$$\frac{ \frac{ 4}{ 3}}{1 } \Rightarrowsin r = \frac{ 3}{4 }$$\Rightarrow$ r = $49^0$$\\ As shown in the figure, \frac{ x}{2.5 } = tan r = 1.15\\ \Rightarrow x = 2.5\times1.15 = 2.8m. 16 17. An object P is focused by a microscope M. A glass slab of thickness 2.1 cm is introduced between P and M. If the refractive index of the slab is 1'5, by what distance should the microscope be shifted to focus the object again ? ##### Solution : The thickness of the glass is d = 2. cm and \mu = 1.5 shift due to the glass slab\\ \Delta T = \big(1 - \frac{1}{\mu}\big)d = \big(1-\frac{1}{1.5}\big)2.1 = 0.7 cm\\ So , the microscope should be shifted 0.70cm o focus the object again. The thickness of the glass is d = 2. cm and \mu = 1.5 shift due to the glass slab\\ \Delta 17 18. A vessel contains water upto a height of 20 cm and above it an oil upto another 20 cm. The refractive indices of the water and the oil are 1'33 and 1'30 respectively. Find the apparent depth of the vessel when viewed from above. ##### Solution : Shift due to water \Delta$$t_w$ = $\big(1 - \frac{1}{\mu}\big) d$ = $\big(1 - \frac{1}{1.33}\big)20$ = 5 cm$\\$ shift due to oil,$\Delta$$t_0 = \big(1 - \frac{1}{1.3}\big) 20 = 4.6 cm\\ Total shift \Deltat = 5 + 4.6 = 9.6 cm\\ Apparent depth = 40 - (9.6) = 30.4 cm below the surface. 18 19. Locate the image of the point P as seen by the eye in the figure (18-E5). ##### Solution : The presence of air medium in between the sheets does not affect the shift .\\ The shift will be due to 3 sheets of diffrent reactive index other than air.\\ = \big(1 - \frac{1}{1.2}\big)(0.2) + \big(1 - \frac{1}{13}\big)(0.3) + \big(1 - \frac{1}{14}\big)(0.4) \\ = 0.2 cm above point p. 19 20. k transparent slabs are arranged one over another. The refractive indices of the slabs are p, p2, p pk and the thicknesses are t1, t2, t3, .......th An object is seen through this combination with nearly perpendicular light. Find the equivalent refractive index of the system which will allow the image to be formed at the same place. ##### Solution : Total number of slabs = k, thickness = t_1,t_2,......t_k \\ Refractive index = \mu_1,\mu_2.......\mu_k \\ Then The shift \Deltat = \big (1-\frac{1}{\mu_1}\big)t_1 + \big (1-\frac{1}{\mu_2}\big)t_2 .......... \big (1-\frac{1}{\mu_k}\big)t_k .......(1) \\ \Deltat = \big (1-\frac{1}{\mu_1}\big)(t_1 + t_2 +....t_k) ........(2) \\ equation 1 and 2 we get \\ \big (1-\frac{1}{\mu_1}\big)(t_1 + t_2 +....t_k) = \big (1-\frac{1}{\mu_1}\big)(t_1 ) +\big (1-\frac{1}{\mu_1}\big)(t_2 )........ \big (1-\frac{1}{\mu_k}\big)(t_k) \\ = (t_1 + t _ 2 + .....+t_k) - \big(\frac{t_1}{\mu_1}\big) + \big(\frac{t_2}{\mu_2}\big) + .........\big(\frac{t_k}{\mu_k}\big) \\ = -\frac{1}{\mu}\sum_{i = 1} ^ {k}t_1 = -\sum_{i = 1} ^ {k} \big(\frac{t_1}{\mu_1}\big)$$\Rightarrow$$\mu = \frac{\sum_{i = 1} ^ {k}t_i }{ \sum_{i = 1} ^ {k}(\frac{t_i}{\mu_i})} 20 21. A cylindrical vessel of diameter 12 cm contains 800E cm 3 of water. A cylindrical glass piece of diameter 8'0 cm and height 8'0 cm is placed in the vessel. If the bottom of the vessel under the glass piece is seen by the paraxial rays (see figure 18-E6), locate its image. The index of refraction of glass is 1'50 and that of water is 1'33. ##### Solution : Given r = 6 cm, r_1 = 4 cm. h_1 = 8 cm. \\ let h = final height of water column. \\ The volume of the cylindrical water column after the glass piece is put will be. \\ \pi r^2h = 800\pi + \pi r_1^2h_1 \\ or r^2h = 800 + r_1^2h_1 \\ or r^2h = 800 + 4^2 \times8 = 25.7 cm \\ there are two shifts due to glass buk as well as water. \\ So,\Delta$$t_1$ = $\big(1-\frac{1}{\mu_0}\big)t_0$ = $\big(1-\frac{1}{\frac{3}{2}}\big)8$ = 2.26 cm $\\$ and $\Delta$$t_2 = \big(1-\frac{1}{\mu_w}\big)t_w = \big(1-\frac{1}{\frac{4}{3}}\big)(25.7 - 8) = 4.44 cm\\ Total shift = (2.66 + 4.44) cm = 7.1 cm above the button. 21 22. Consider the situation in figure (18-E7). The bottom of the pot is a reflecting plane mirror, S is a small fish and T is a human eye. Refractive index of water is (a) At what distance(s) from itself will the fish see the image(s) of the eye ? (b) At what distance(s) from itself will the eye see the image(s) of the fish. ##### Solution : Let x = distance of the image of the eye formed above the surface as seen by the fish.\\ So, \frac{H}{x} = \frac{ Real depth}{ apparent depth} = \frac{ 1}{\mu } \\ So, distance of the direct image = \frac{ H}{2 } + \mu H = H(\mu + \frac{1 }{2 })$$\\$ Similarly , image through mirror = $\frac{ H}{2 }$ + (H + x) = $\frac{3H}{2}$ +$\mu$H = $\mu(\mu + \frac{3}{2})$ $\\$ B) Here $\frac{\frac{H}{2}}{y}$ = \mu , So. y = $\frac{\mu}{2\mu}$$\\Where, y = distance of the image of fish below the surface as seen by eye.\\ So, the direct image = H + y = \mu + \frac{H}{2\mu} = H\big(1 + \frac{1}{2\mu}\big) \\ again another image of fish will be formed \frac{H}{2} below the mirror.\\ So, the real depth for that image of fish becomes H + \frac{H}{2} = \frac{3H}{2}$$\\$ So.the apparent depth from the surface of water = $\frac{3\mu}{2\mu}$ $\\$ So. distance of the image from the eye = H + $\frac{\frac{3H}{2}}{2\mu}$ = H$(1 + \frac{3}{2\mu})$$\\ 22 23. A small object is placed at the centre of the bottom of a cylindrical vessel of radius 3 cm and height 4 cm filled completely with water. Consider the ray leaving the vessel through a corner. Suppose this ray and the ray along the axis of the vessel are used to trace the image. Find the apparent depth of the image and the ratio of real depth to the apparent depth under the assumptions taken. Refractive index of water = 113. ##### Solution : According to the figure ,\frac{x}{3} = cot r ....(1)\\ Again \frac{sin i }{sinr } = \frac{ 1}{1.33 } = \frac{ 3}{ 4}$$\\$$\Rightarrow sin r = \frac{4 }{3 } sin i = \frac{ 4}{3 }\times \frac{ 3}{ 5} = \frac{ 4}{ 5}$$\\$ $\Rightarrow$ cot r = $\frac{ 3}{ 4}$..........(2)$\\$ from (1) and (2) = $\frac{ x}{3 }$ = $\frac{ 3}{4 }$ $\\$ then x = $\frac{ 9}{ 4}$ = 2.25cm.$\\$ $\therefore$ Ratio of real and apparent depth = 4 : (2.25) = 1.78.

23   24. A cylindrical vessel, whose diameter and height both are equal to $30 cm$, is placed on a horizontal surface and a small particle $P$ is placed in it at a distance of $5.0 cm$ from the centre. An eye is placed at a position such that the edge of the bottom is just visible (see figure 18-E8). The particle $P$ is in the plane of drawing. Up to what minimum height should water be poured in the vessel to make the particle $P$ visible ? ##### Solution : For the given cylindrical vessel,diameter = 30 cm.$\\$ then r = 15 cm and h = 30 cm $\\$ Now, $\frac{ sin i}{ sin r}$ = $\frac{ 3}{ 4}\big[\mu_w = 1.33 = \frac{ 4}{3 }\big]$ $\\$ then sin i = $\frac{ 3}{ 4\sqrt2}$ [because r = $45^0$] $\\$ The point p will be visible when the refracted ray makes angle $45^0$ at pont refraction, $\\$ Let x = distance of point p from x. $\\$ Now tan $45^0$ = $\frac{ x + 10 }{d }$ $\\$ then d = x + 10.........(1) $\\$ again tan i = $\frac{ x}{d }$ $\\$ $\Rightarrow\frac{3 }{ \sqrt{23}} = \frac{d - 10 }{ d}$ $\\$ since sin i = $\frac{3 }{4\sqrt2 }$$\\ and tan i = \frac{3 }{2\sqrt{23}} \\ \Rightarrow \frac{3 }{ \sqrt{23}} - 1 = - \frac{10 }{d } \\ d = \frac{\sqrt{23}\times10 }{ \sqrt{23}\times - 3} = 26.7 cm. 24 25. A light ray is incident at an angle of 45° with the normal to a 12 cm thick plate (1 = 2.0). Find the shift in the path of the light as it emerges out from the plate. ##### Solution : As shown in the figure,\\ \frac{sin 45^0 }{ sin r} = \frac{ 2}{ 1} \\ then sin r = \frac{ sin 45^0}{ 2} = \frac{ 1}{ 2\sqrt2}$$\\$ r = $21^0$ $\\$ Therefore, $\Theta$ = $(45^0 - 21^0)$ = $24^0$ $\\$ Here BD = shift in path = AB sin $24^0$ $\\$ = 0.406 x AB = $\frac{ AE}{cos 21^0 }$ x 0.406 = 0.62 cm.

25   26. An optical fibre $(p = 1.72)$ is surrounded by a glass coating $(II = 1'50)$. Find the critical angle for total internal /reflection at the fibre-glass interface.

##### Solution :

For calculation of critical angle,$\\$ $\frac{sin i }{sin r }$ = $\frac{ \mu_2}{ \mu_1}$$\Rightarrow \frac{ sin C}{sin 90 } = \frac{15 }{ 1.72} = \frac{ 75}{86 }$$\\$ $\Rightarrow$C = $sin^{-1}\big(\frac{75}{26}\big)$

26   27. A light ray is incident normally on the face $AB$ of a right-angled prism ABC (\mu= P50) as shown in figure (18-E9). What is the largest angle 4) for which the light ray is totally reflected at the surface $AC$ ?

##### Solution :

Let $\Theta_c$ be the critical angle for the glass, $\\$ $\frac{sin \Theta_c }{sin 90^0 }$ = $\frac{1 }{x }$$\Rightarrowsin\Theta_c = \frac{ 1}{1.5 } = \frac{ 2}{3 } \Rightarrow$$\Theta_c$ = $sin^{-1}\big(\frac{2}{3}\big)$ $\\$ From figure for total internal reflection $90^0 - \phi > \Theta_c$ $\\$ then $\phi<90^0 - \Theta_c$ $\\$ $\Rightarrow$ $\phi < cos^{-1}(\frac{2}{3})$ $\\$ So, the largest angle for which is totally reflected at the surface is $cos^{-1}(\frac{2}{3})$

27   28. Find the maximum angle of refraction when a light ray is refracted from glass $(GI = 1'50)$ to air.

##### Solution :

From the definition of the criticle angle,if refracted angle is more than $90^0$, then reflection occurs,which is known as total internal reflection.$\\$ So,maximum angle of reflection is $90^0$.

28   29. Light is incident from glass $(11 = 1'5)$ to air. Sketch the variation of the angle of deviation 5 with the angle of incident i for$0 < i < 90°$.

##### Solution : Refractive index of glass $\mu_g= 1.5$$\\ Given 0^0 < i < 90^0 \\ Let C = criticle angle \\ \frac{ sin C}{sin r } = \frac{\mu_a }{\mu_g }\Rightarrow \frac{sin C }{sin 90^0 } = \frac{ 1}{15 } = 0.66 \\ then C = 40^048' '$$\\$ the angle of deviation due to refraction from glass to air increases as the angle of incidence increases from $0^0$ to $40^048''.$ the angle deviation due to total internal reflection further increases $40^048''.$ to $45^0$ and then it decreases.

29   30. Light is incident from glass $(OA = 1'50)$ to water $(11 = P33)$. Find the range of the angle of deviation for which there are two angles of incidence.

##### Solution : $\mu_g$ = 1.5 = $\frac{ 3}{2 }$;$\mu_w$ = 1.33 = $\frac{ 4}{ 3}$$\\ for two angle of incidence.\\ 1) when light passes straight through noral,\\ angle of incidence = 0^0,angle of regfraction = 0^0, angle of deviation = 0, \\ 2) when light is incident at criticle angle ,\\ \frac{ sin C}{ sin r} = \frac{ \mu_w}{ \mu_g} (since light passing from glass to water)\\ sinC = \frac{8 }{ 9}$$\\$ C = $sin^{-1}(\frac{8}{9})$ = $62.73^0$$\\ \therefore Angle of deviation = 90^0 - C = 90^0 - sin^{-1}(\frac{8}{9}) = cos^{-1}(\frac{8}{9}) = 37.27^0.\\ Here, if the angle of incidence is increased beyond critical angle, total internal reflection occurs and deviation decreases. So, the range of deviation is 0 to cos^{-1}(\frac{8}{9}) 30 31. Light falls from glass (.1 = P5) to air. Find the angle of incidence for which the angle of deviation is 90^0. ##### Solution : Since , \mu = 1.5, Criticle angle = sin^{-1}(\frac{1}{\mu}) = sin^{-1}(\frac{1}{1.5}) = 41.8^0$$\\$ We know, the maximum attainable deviation in refraction is $(90^0 - 41.8^0)$ = $47.2^0$ So, in this case, total internal reflection must have taken place.$\\$ In reflection, Deviation = $180^0$ - 2i = $90^0$ $\\$ then i = $45^0$$\\ 31 32. A point source is placed at a depth h below the surface of water (refractive index =\mu). (a) Show that light escapes through a circular area on the water surface with its centre directly above the point source. (b) Find the angle subtended by a radius of the area on the ##### Solution : 32 32. A point source is placed at a depth h below the surface of water (refractive index =\mu). (a) Show that light escapes through a circular area on the water surface with its centre directly above the point source. (b) Find the angle subtended by a radius of the area on the ##### Solution : Let x = radius of the circular area\\ \frac{ x}{h } = tan C(where C is criticle angle) \\ \frac{ x}{h } = \frac{ sin C}{\sqrt{1-sin^2C }} = \frac{ \frac{1}{\mu}}{\sqrt{1-\frac{1}{\mu^2}} } \\ \frac{ x}{h } = \frac{ 1}{\sqrt{\mu^2 - 1 }} or x = \frac{ h}{\sqrt{\mu^2 - 1 }} \\ So, light escapes through a circular area on the water surface directly above the point source.\\ b) Angle subtained by a radius of the area on the source, C = sin^{-1}(\frac{1}{\mu}). 33 33. A container contains water upto a height of 20 cm and there is a point source at the centre of the bottom of the container. A rubber ring of radius r floats centrally on the water. The ceiling of the room is 2.0 m above the water surface. (a) Find the radius of the shadow of the ring formed on the ceiling if r = 15 cm. (b) Find the maximum value 'of r for which the shadow of the ring is formed on the ceiling. Refractive index of water = \frac{4}{3} ##### Solution : a) As shown in the fig sin i = \frac{15 }{25 }$$\\$ So,$\frac{ sin i}{ sin r}$ = $\frac{1 }{\mu }$ = $\frac{3 }{4 }$ $\\$ sin r = $\frac{ 4}{5 }$ $\\$ Again $\frac{ x}{ 2}$ = tan r(from figure) $\\$ So,sin r = $\frac{tan r }{ \sqrt{1 + tan^2 }}$ = $\frac{\frac{x}{2}}{\sqrt{1 - \frac{x^2}{4}}}$ $\\$ = $\frac{x}{ \sqrt{4 + x^2 }}$ = $\frac{4}{5}$ $\\$ $\Rightarrow$ 25 $x^2$ = 16$(4 + x^2)$ $\Rightarrow$ 9$x^2$ $\Rightarrow$x = $\frac{8 }{3 }$m. $\\$ $\therefore$ total radius of shadow = $\frac{ 8}{3 }$ + 0.15 = 2.81 m. $\\$ b) for maximum size of the ring i = criticle angle = C $\\$ let r = maximum radius $\\$ sin C = $\frac{ sin C}{sin R }$ = $\frac{R }{\sqrt{20^2 + R^2} }$ = $\frac{ 3}{ 4}$ (since sin r = 1) $\\$ 16$R^2$ = 9$R^2$ + 9 x 400 $\\$ 7$R^2$ = 9 x 400 $\\$ R = 22.67 cm,

34   34. Find the angle of minimum deviation for an equilateral prism made of a material of refractive index $P732$. What is the angle of incidence for this deviation ?

##### Solution : given A = $60^0$. $\mu$ = 1.732$\\$ since angle of minimum deviation is given by $\\$ $\mu$ = $\frac{ sin\big(\frac{A + \delta\mu}{2}\big)}{sin\frac{A}{2} }$$\\ 1.732 x \frac{1}{2} = sin(30 +\frac{\delta\mu}{2}) \\ sin^{-1}(0.866) = 30 + \frac{\delta\mu}{2} = 60^0 = 30^0\frac{\delta\mu}{2}$$\\$ $\delta\mu$ = $60^0$$\\ Now. \delta\mu = i + i^1 - A\\ \Rightarrow60^0 = i + i^1 - 60^0$$\\$ $\Rightarrow$ i = $60^0$.So. is the angle of incidence must be $60^0$

35   35. Find the angle of deviation suffered by the light ray shown in figure $(18-E10)$. The refractive index $µ = P5$ for the prism material.

##### Solution : given $\mu$ = 1.5$\\$ and angle of prism = $4^0$$\\ \therefore\mu = \frac{sin\big(\frac{A + \delta_m}{2}\big)}{sin\frac{A}{2}} =\frac {\frac{(A + \delta_m)}{2}}{(\frac{A}{2})}$$\\$ $\therefore\mu$ = $\frac{A + \delta_m}{2}$ $\Rightarrow$ 1.5 = $\frac{4^0 + \delta_m}{4^0}$$\\ \delta_m = 4^0 x (1.5) - 4^0 = 2^0 36 36. A light ray, going through a prism with the angle of prism 60°, is found to deviate by 30°. What limit on the refractive index can be put from these data ? ##### Solution : Given A = 60^0 and \delta = 30^0$$\\$ We know that $\\$ $\mu$ = $\frac{sin\big(\frac{A + \delta_m}{2}\big)}{sin \frac{A}{2}}$ = $\frac{sin\big(\frac{60^0+ \delta_m}{2}\big)}{sin 30^0}$ = 2 sin$\frac{60^0 + \delta_m}{2}$$\\ Since, one ray has been found out which has deviated by 30^0, the angle of minimum deviation should be either equal or less than 30^0. (It can not be more than 30^0).\\ So ,\mu\le2sin\frac{60^0 + \delta_m}{2}$$\\$ or $\mu\le2\times\frac{1}{\sqrt2}$ $\\$

37   37. Locate the image formed by refraction in the situation shown in figure $(18-E11)$.

##### Solution : $\mu_1$ = 1,$\mu_2$ = 1.5, R = 20 cm(radius of curvature), u = -25 cm,$\\$ $\therefore\frac{\mu_2}{v} - \frac{\mu_1}{u}$ = $\frac{\mu_2 -\mu_1 }{R}$ $\\$ then $\frac{1.5}{v}$ = $\frac{0.5 }{20 }$ - $\frac{1 }{25 }$ = $\frac{1 }{40 }$ - $\frac{1 }{25 }$ =$\frac{-3}{200}$ $\\$ v = -200 x 0.5 = -100 cm. $\\$ So, the image is 100 cm from (p) the surface on side of S.

38   38. A spherical surface of radius $30 cm$ separates two transparent media $A$ and' $B$ with refractive indices $1'33$ and $P48$ respectively. The medium A is on the convex side of the surface. Where should a point object be placed in medium A so that the paraxial rays become parallel after refraction at the surface ?

##### Solution :

Since, paraxial rays become parallel after refraction i.e. image is formed at $\infty$. $\\$ v = $\infty$,$\mu_1$ = 1.33. u = ?,$\mu_2$ = 1.48. R = 30 cm. $\\$ $\frac{\mu_2 }{V} - \frac{\mu_1}{u}$ =$\frac{\mu_2 - \mu_1}{R}$ $\\$ then $\frac{1.48 }{\infty}$ - $\frac{1.33 }{u}$ =$\frac{1.48 - 1.33 }{30}$ $\\$ then -$\frac{1.33 }{u}$ - $\frac{0.15 }{30}$ $\\$ then u = -266.0 cm $\\$ Object should be placed at a distance of 266 cm from surface (convex) on side A.

39   39. Figure $(18-E12)$ shows a transparent hemisphere of radius $3.0 cm$ made of a material of refractive index $2-0$. (a) A narrow beam of parallel rays is incident on the hemisphere as shown in the figure. Are the rays totally reflected at the plane surface ? (b) Find the image formed by the refraction at the first surface. (c) Find the image formed by the reflection or by the refraction at the plane surface. (d) Trace qualitatively the final rays as they come out of the hemisphere.

Given ,$\mu_2$ = 2.0 $\\$ So, criticle angle = $sin^{-1}\big(\frac{1}{\mu_2}\big)$ = $sin^{-1}\big(\frac{1}{2}\big)$ = $30^0$$\\ a) As angle of incidence is greater than the critical angle, the rays are totally reflected internally. \\ b) Here \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$\\$ $\frac{2 }{v }$ - $\big(-\frac{1 }{\infty }\big)$ = $\frac{2 - 1 }{3 }$$\\ \frac{2 }{v } = \frac{1 }{ 3} \\ v = 6cm.\\ If the sphere is completed, image is formed diametrically opposite of A.\\ c) Image is formed at the mirror in front of A by internal reflection. 40 40. A small object is embedded in a glass sphere (p = 1'5) of radius 5.0 cm at a distance 1'5 cm left to the centre. Locate the image of the object as seen by an observer standing (a) to the left of the sphere and (b) to the right of the sphere. ##### Solution : a) image seen from left : \\ u = (5 - 15) = -3.5 cm.\\ R = -5 cm.\\ \therefore \frac{\mu_2 }{v } - \frac{\mu_1 }{u } = \frac{\mu_2- \mu_1 }{R } \\ then \frac{1}{v } + \frac{1.5 }{3.5} = -\frac{1 - 1.5 }{5 }$$\\$ then $\frac{1 }{v }$ = $\frac{1}{10}$ - $\frac{3 }{7 }$$\\ v = \frac{-70 }{23} = -3 cm (inside the sphere)\\ image will be formed 2 cm left to centre. 41 41. A biconvex thick lens is constructed with glass (p = 1'50). Each of the surfaces has a radius of 10 cm and the thickness at the middle is 5 cm. Locate the image of an object placed far away from the lens. ##### Solution : R_1 = R_2 =10 cm. t = 5 cm. u = -\infty \\ for the first reaction (at A)\\ \frac{\mu_g }{v } - \frac{\mu_a }{u } = \frac{\mu_g - \mu_a }{R_1} or \frac{1.5}{v } - 0 = \frac{1.5}{10 }$$\\$ v = 30 cm.$\\$ again fo second surface u = (30 - 5) = 25 cm (virtual object)$\\$ $R_2$ = -10 cm.$\\$ So, $\frac{1}{v }$ - $\frac{15 }{25 }$ = $\frac{-0.5}{-10 }$$\\ v = 9.1 cm.\\ So, the image is formed 9.1 cm further from the second surface of the lens. 42 42. A narrow pencil of parallel light is incident normally on a- solid transparent sphere of radius r. What should be the refractive index if the pencil is to he focused (a) at the surface of the sphere, (b) at the centre of the sphere. ##### Solution : For the reaction at convex surface A\\ \mu = -\infty,\mu = 1,\mu = ?.\\ a) when focused on the surface v = 2r,R = r\\ So, \frac{ \mu_2}{v } - \frac{ \mu_1}{u} = \frac{ \mu_2 - \mu_1}{R}$$\\$ then $\frac{ \mu_2}{2r }$ = $\frac{ \mu_2 - 1}{r }$$\\ then \mu_2 = 2\mu_2 - 2 \Rightarrow$$\mu_2$ = 2.$\\$ b) when focused at centre, u = r, R = r$\\$ So, $\frac{ \mu_2}{v }$ - $\frac{ \mu_1}{u}$ = $\frac{ \mu_2 - \mu_1}{R}$ $\\$ $\frac{ \mu_2}{R }$ = $\frac{ \mu_2 - 1}{r }$$\\ \mu_2 = \mu_2 - 1 \\This is not possible.\\ So, it cannot focus at the centre. 43 43. One end of a cylindrical glass rod (p = 1'5) of radius 1.0 cm is rounded in the shape of a hemisphere. The rod is immersed in water (p = 4/3) and an object is placed in the water along the axis of the rod at a distance of 8'0 cm from the rounded edge. Locate the image of the object. ##### Solution : Radius of the cylindrical glass tube = 1 cm\\ we know,\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$\\$ Here u = - 8 cm, $\\$ $\mu_2$ = $\frac{3}{2}$, $\\$ $\mu_1$ = $\frac{4}{3}$,$\\$ R = +1 cm, $\\$ So,$\frac{3 }{ 2v}$ +$\frac{4 }{ 3\times8}$ $\Rightarrow$$\frac{3 }{ 2v} + \frac{1 }{ 6} = \frac{1 }{ 6} , v = \infty,\\ \therefore the image will be formed at infinity. 44 44. A paperweight in the form of a hemisphere of radius 3'0 cm is used to hold down a printed page. An observer looks at the page vertically through-the paperweight. At what height above the page will the printed letters near the centre appear to the observer ? ##### Solution : in the first refraction at A.\\ \mu_2 = \frac{3}{2},\mu_1 = 1, u = 0, R = \infty,\\ So,\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$\\$ then v = 0 since (R = $\infty$ and u = 0)$\\$ $\therefore$ the image will be formed at the point,now for the second refraction at B,$\\$ u = -3 cm, R = -3 cm,$\mu_1$ = $\frac{3}{2}$,$\mu_2$ = 1,$\\$ So, $\frac{1}{v}$ + $\frac{3}{2\times3}$ = $\frac{1 - 1.5}{-3}$ = $\frac{1}{6}$$\\ \frac{1}{v} = \frac{1}{6 } - \frac{1}{2} = -\frac{1}{3}$$\\$ v = -3 cm,$\\$ there will be no shift in the final image.

45   45. Solve the previous problem if the paperweight is inverted at its place so that the spherical surface touches the paper.

##### Solution : Thickness of glass = 3 cm, $\mu_g$ = 1.5$\\$ image shift = 3$\big( 1 -\frac{1}{1.5}\big)$ $\\$ [Treating it as a simple refraction problem because the upper surface is flat and the spherical surface is in contact with the object]$\\$ = 3 x $\frac{0.5}{1.5}$ = 1 cm$\\$ The image will appear 1 cm above the point p,

46   46. A hemispherical portion of the surface of a solid glass sphere $(p = 1•5)$ of radius r is silvered to make the inner side reflecting. An object is placed on the axis of the hemisphere at a distance $3r$ from the centre of the sphere. The light from the object is refracted at the unsilvered part, then reflected from the silvered part and again refracted- at the unsilvered part. Locate the final image formed.

##### Solution : As shown in the figure OQ = 3r, OP = r,PQ = 2r$\\$ For refraction at APB$\\$ we know $\frac{\mu_2}{v}$ - $\frac{\mu_1}{u}$ = $\frac{\mu_2 - \mu_1}{R}$$\\ \frac{1.5}{v} - \frac{1}{2r} = \frac{0.5}{r} = \frac{1}{2r}$$\\$ v = $\infty$ $\\$ for the reflection in concave mirror $\\$ Here u = $\frac{-3r}{2}$$\\ \frac{1}{v} - \frac{1.5}{\frac{-3r}{2}} = \frac{-3r}{2} = \frac{-0.5}{r}$$\\$ [Here $\mu_1$ = 1.5 and $\mu_2$ = 1 and R = -r]$\\$ v = -2r$\\$ As, negative sign indicates images are formed inside APB. So, image should be at C. So, the final image is formed on the reflecting surface of the sphere.

47   47. The convex surface of a thin concavo-convex lens of glass of refractive index $1.5$ has a radius of curvature $20 cm$. The concave surface has a radius of curvature $60 cm$. The convex side is silvered and placed on a horizontal surface as shown in figure $(18-E13)$. (a) Where should a pin be placed on the axis so that its image is formed at the same place ? (b) If the concave part is filled with water $(p = 4/3)$, find the distance through which the pin should be moved so that the image of the pin again coincides with the pin.

##### Solution : a) let the pin is at a distance of x from the lens.$\\$ then for first reaction $\frac{\mu_2}{v}$ - $\frac{\mu_1}{u}$ = $\frac{\mu_2 - \mu_1}{R}$$\\ Here \mu_2 = 1.5,\mu_1 = 1, u = - x, R = -60 cm,\\ \therefore \frac{1.5}{v} - \frac{1}{-x} = \frac{0.5}{-60}$$\\$ $\Rightarrow$ 120(1.5x + v) = -vx$\\$ $\Rightarrow$(120 + x) = -180x$\\$ $\Rightarrow$ v = $\frac{180x}{120 + x}$$\\ this image distance agin object distance for the concave mirror. u = \frac{-180x }{120 + x }, f= -10 cm,(\therefore f =\frac{R}{2}) \\ \therefore \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \frac{1}{v_1} = \frac{1}{-10} - \frac{120 + x }{180x}$$\\$ $\frac{1}{v_1}$ = $\frac{120 + x -18x}{180x}$$\\ v_1 = \frac{180x}{120 - 17x}$$\\$ Again the image formed is refracted through the lens so that the image is formed on the object taken in the 1st refraction. So, for 2nd refraction.$\\$ Acording to sign conversion v = -x, $\mu_2$ = 1, $\mu_1$ = 1.5, R = -60$\\$ Now $\frac{\mu_2}{v}$ - $\frac{\mu_1}{u}$ = $\frac{\mu_2 - \mu_1}{R}$$\\ \Rightarrow \frac{1}{-x} - \frac{1.5}{180x}(120 - 17x) = \frac{-0.5}{-60}$$\\$ Multiplying both sides with 120 m.we get $\\$ 120 + 120 - 17x = -x$\\$ 16x = 240 $\\$ x = 15 cm.$\\$ Object should be placed at 15 cm from the lens on the axis.

48   48. A double convex lens has focal length $25 cm$. The radius of curvature of one of the surfaces is double of the other. Find the radii, if the refractive index of the material of the lens is $1.5$.

##### Solution : for the convex lens $\\$ f = 25 cm,$R_1$ = R and $R_2$ = -2R $\\$ $\frac{1}{f}$ =$(\mu - 1 )\big(\frac{1}{R_1} - \frac{1}{R_2}\big)$ $\\$ $\frac{1}{25}$ = $\frac{3}{4}\frac{1}{R}$ $\\$ then R = 18.75 cm. $\\$ $R_1$ = 18.75 cm, $\\$ $R_2$ = 2R = 37.5 cm$\\$

for the convex lens$\\$ f = 25 cm,$R_1$ = R and $R_2$ = -2R$\\$ $\frac{1}{f}$ = ($\mu$ - 1 )$\big($\frac{1}{R_1} - $\frac{1}{R_2}\big)$$\\ \frac{1}{25} = \frac{3}{4} \frac{1}{R}$$\\$ then R = 18.75 cm.$\\$ $R_1$ = 18.75 cm, $R_2$ = 2R = 37.5 cm$\\$

49   49. The radii of curvature of a lens are $+ 20 cm$ and $+ 30 cm$. The material of the lens has a refracting index $1'6$. Find the focal length of the lens (a) if it is placed in air, and (b) if it is placed in water $(p = 1'33)$.

##### Solution : $R_1$ = 20 cm; $R_2$ = 30 cm.$\mu$ = 1.6$\\$ a) if placed in air : $\frac{1}{f}$ = $(\mu_0 - 1)\big(\frac{1}{R_1} - \frac{1}{R_2}\big)$ = $\big(\frac{1.6}{1} - 1\big)\big(\frac{1}{20} - \frac{1}{30}\big)$ $\\$ f = $\frac{60}{6}$ = 100 cm.$\\$ b) if placed in water :$\\$ $\frac{1}{f}$ = $(\mu_w - 1)\big(\frac{1}{R_1} - \frac{1}{R_2}\big)$ =$\big(\frac{1.6}{1.33} - 1\big)\big(\frac{1}{20} - \frac{1}{30}\big)$ $\\$ $\Rightarrow$ f = 300 cm.

50   50. Lenses are constructed by a material of refractive index $1'50$. The magnitude of the radii of curvature are $20 cm$ and $30 cm$. Find the focal lengths of the possible lenses with the above specifications.

##### Solution :

Given $\mu$ = 1.5 $\\$ Magnitude of radii curvature = 20 cm and 30 cm $\\$ the 4 types of possible lens are below. $\\$ $\frac{1}{f}$ = $(\mu - 1)\big(\frac{1}{R_1} - \frac{1}{R_2}$ $\\$ case(1): (double convex ) $[R_1= +ve,R_2 = -ve]$ $\\$ $\frac{1}{f}$ =(15 - 1)$\big(\frac{1}{20} - \frac{1}{30}\big)\Rightarrow f = -24 cm$ $\\$ case(2): (double concave )$[R_1 = -ve,R_2 = +ve]$ $\\$ $\frac{1}{f}$ =(15 - 1)$\big(\frac{-1}{20} - \frac{1}{30}\big)\Rightarrow f = -24 cm$ $\\$ case(3): (concave concave )$[R_1 = -ve,R_2 = +ve]$ $\\$ $\frac{1}{f}$ =(15 - 1)$\big(\frac{1}{20} - \frac{1}{30}\big)\Rightarrow f = 120 cm$ $\\$ 51   51. A thin lens made of a material of refractive index $p2$ has a medium of refractive index $p$, on one side and a medium of refractive index $p$, on the other side. The lens is biconvex and the two radii of curvature have equal magnitude R. A beam of light travelling parallel to the principal axis is incident on the lens. Where will the image be formed if the beam is incident from (a) the medium $p$, and (b) from the medium $p3$ ?

##### Solution : when the beam is incident on the lens from medium $\mu_1$. $\\$ then $\frac{\mu_2}{v} - \frac{\mu_2}{u}$ = $\frac{\mu_2 - \mu_1}{R} or \frac{\mu_2}{v} - \frac{\mu_1}{(-\infty)}$ = $\frac{\mu_2 - \mu_1}{R}$ $\\$ or $\frac{1}{v}$ = $\frac{\mu_2 - \mu_1}{R} or v = \frac{\mu_2R}{\mu_2 - \mu_1}$ $\\$ again for second refraction $\\$ $\frac{\mu_3}{v} - \frac{\mu_2}{u}$ = $\frac{\mu_3 - \mu_2}{R}$ $\\$ or $\frac{\mu_3}{v}$ = - $\big[\frac{\mu_3- \mu_2}{R} - \frac{\mu_2}{\mu_2R}(\mu_2 - \mu_1)\big]$ $\\$ then - $\big[\frac{\mu_3 - \mu_2 -\mu_2 + \mu_1}{R}\big]$ $\\$ v = - $\big[\frac{\mu_3R}{\mu_3 - 2\mu_2 + \mu_1}\big]$ $\\$ so the image will be formed at = $\big[\frac{\mu_3R}{2\mu_2 - \mu_1-\mu_3}\big]$ $\\$ B) similarly for the beam from \mu_3 medium the image is formed at $\big[\frac{\mu_1R}{2\mu_2 - \mu_1-\mu_3}\big]$ 52   52. A convex lens has a focal length of $10 cm$. Find the location and nature of the image if a point object is placed on the principal axis at a distance of $(a) 9.8 cm$, $(b) 10'2 cm$ from the lens.

##### Solution : given that f = 10 cm,$\\$ a) when u = -9.5 cm$\\$ $\frac{1}{v} - \frac{1}{u}$ = $\frac{1}{f}\Rightarrow$ $\\$ $\frac{1}{v} = \frac{1}{10} - \frac{1}{9.8}$= $\frac{-0.2}{9.8}$ $\\$ v = -490 cm,$\\$ so m = $\frac{v}{u} = \frac{-490}{-9.8}= 50 cm$ $\\$ so the image is erect and virtual$\\$ b)when u = -10.2 cm $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\Rightarrow$ $\\$ $\frac{1}{v} = \frac{1}{10} - \frac{1}{-10.2} = \frac{102}{0.2}$ $\\$ v = 510 cm$\\$ so m = $\frac{v}{u} = \frac{510}{-9.8}$ $\\$ the image is real and inverted

53   53. A slide projector has to project a $35 mm$ slide $(35 mm \times 23 mm)$ on a $2m \times 2m$ screen at a distance of 10 m from the lens. What should be the focal length of the lens in the projector ?

##### Solution :

for the projector the magnification required is given by$\\$ m = $\frac{v}{u}$ =$\frac{200}{3.5}$$\\ then u = 17.5 cm\\ [35mm > 23 mm so the magnification is calculated taking object size 35 mm]\\ \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$\\$ $\frac{1}{v} -\frac{1}{-u} = \frac{1}{f}$$\\ \frac{1}{1000} + \frac{1}{17.5} = \frac{1}{f}$$\\$ f = 17.19 cm.

54   54. A particle executes a simple harmonic motion of amplitude $1.0 cm$ along the principal axis of a convex lens of focal length $12 cm$. The mean position of oscillation is at $20 cm$ from the lens. Find the amplitude of oscillation of the image of the particle.

##### Solution : when the object is at 19 cm from the lens let the image will be at $v_1$$\\ \frac{1}{v_1} - \frac{1}{u_1} =\frac{1}{f} \\ \frac{1}{v_1} - \frac{1}{-19} = \frac{1}{12} \\ v_1 = 32.57cm \\ again when the object is at 21 cm from the lens .let the image will be at v_1 \\ \frac{1}{v_1} -\frac{1}{u_2} = \frac{1}{f}$$\\$ $\frac{1}{v_2} + \frac{1}{21} = \frac{1}{12}$$\\ v_2 = 28 cm.\\ amplitude of vibration of the image is A = \frac{32.57 - 28}{2} = 2.285 cm 55 55. An extended object is placed at a distance of 5'0 cm from a convex lens of focal length 8'0 cm. (a) Draw the ray diagram (to the scale) to locate the image and from this, measure the distance of the image from the lens. (b) Find the position of the image from the lens formula and see how close the drawing is to the correct result. ##### Solution : given u = -5 cm,f = 8 cm\\ so \frac{1}{v}- \frac{1}{u} = \frac{1}{f}$$\\$ then $\frac{1}{8}- \frac{1}{5} = \frac{-3}{40}$$\\ v = -13.3 cm 56 56. A pin of length 2'00 cm is placed perpendicular to the principal axis of a converging lens. An inverted image of size 1-00 cm is formed at a distance of 40'0 cm from the pin. Find the focal length of the lens and its distance from the pin. ##### Solution : given that (-u) + v = 40 cm = distance between object and image \\ h_0 = 2 cm,h_1= 1cm \\ since \frac{h_1}{h_0} = \frac{v}{-u} = magnification \\ \frac{1}{2} = \frac{v}{-u} \\ u = -2v \\ now \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \\ \frac{1}{v} +\frac{1}{2v} = \frac{1}{f} \\ \frac{3}{2v} = \frac{1}{f} \\ f = \frac{2v}{3} \\ again (-u) + v = 40\\ 3v = 40\\ v = \frac{40}{3}cm \\ f = \frac{2\times40}{3\times3} =8.89 cm = foval length \\ from 1 and 2 \\ u = -2v = -3f = -3(8.89) = 26.7 cm = object distance 57 57. A convex lens produces a double size real image when an object is placed at a distance of 18 cm from it. Where should the object be placed to produce a triple size real image ? ##### Solution : a real image is formed so magnification m = -2 (inverted image)\\ \frac{v}{u} = -2 \\ v = -2u = (-2)(-18) = 36 \\ from lens formula \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \\ \frac{1}{36} - \frac{1}{-18} =\frac{1}{f} \\ f = 12 cm\\ now for triple sized image = m = -3 = \frac{v}{u}$$\\$ $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$\\ \frac{1}{v} - \frac{1}{u} =\frac{1}{f}$$\\$ $\frac{1}{-3u} - \frac{1}{u} = \frac{1}{12}$$\\ 3u = -48\\ then u = -16 cm\\ so the object should be placed 16 cm from lens 58 58. A pin of length 2.0 cm lies along the principal axis of a converging lens, the centre being at a distance of 11 cm from the lens. The focal length of the lens is 6 cm. Find the size of the image ##### Solution : Now we have to calculate the image of A and B. Let the images be A_1 and B_1. So, length of A_1 B_1= size of image. \\ for A , u = -10 cm,f = 6 cm \\ since \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \\ \frac{1}{v} - \frac{1}{-10} = \frac{1}{6}$$\\$ v = 15 cm = O$A_1$ $\\$ for B,u = -12cm,f = 6 cm$\\$ again $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $\\$ $\frac{1}{v} = \frac{1}{6} - \frac{1}{12}$ $\\$ v = 12 cm = O$B_1$ $\\$ $A_1B_1 = OA_1-O B_1$ = 15 - 12 = 3 cm $\\$ so size of image = 3 cm

59   59. The diameter of the sun is $1.4 \times 10 .9 m$ and its distance from the earth is $1.5 \times 10 u m$. Find the radius of the image of the sun formed by a lens of focal length $20 cm$.

##### Solution :

60   59. The diameter of the sun is $1.4 \times 10 .9 m$ and its distance from the earth is $1.5 \times 10 u m$. Find the radius of the image of the sun formed by a lens of focal length $20 cm$.

u = $-1.5\times10^{11} m ; f = 20 \times10^{-2}m$$\\ since f is very small compared to u,distance is taken as \infty,so imge will be formed at focus\\ v = 20\times10^{-2}m$$\\$ we know m = $\frac{v}{u} = \frac{h_image}{h_object}$$\\ \frac{20\times10^{-2}}{1.5\times10^{11}} = \frac{D_{image}}{2} = 0.93 mm 61 60. A 5.0 diopter lens forms a virtual image which is 4 times the object placed perpendicularly on the principal axis of the lens. Find the distance of the object from the lens. ##### Solution : given P = 5 diopter(convex lens) \\ f = \frac{1}{5} m = 20 cm \\ since image is formed u and v bothare negative given \frac{v}{u}=4 \\ v = 4u; \\ from lens formula \frac{1}{v} -\frac{1}{u} = \frac{1}{f} \\ \frac{1}{f} = \frac{1}{4u} - \frac{1}{u} \\ then \frac{1}{20} = \frac{1 - 4}{4u} = \frac{-3}{4u} \\ Object is placed 15 cm away from the lens 62 61. A diverging lens of focal length 20 cm and a converging mirror of focal length 10 cm are placed coaxially at a separation of 5 cm. Where should an object be placed so that a real image is formed at the object itself? ##### Solution : Let the object to placed at a distance x from the lens further away from the mirror. For the concave lens (1st refraction) \\ u = -x,f = -20 cm,fro lens formula \\ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \\ \frac{1}{v} = \frac{1}{-20} + \frac{1}{x} \\ v = -\big(\frac{20x}{x + 20}\big) \\ So, the virtual image due to fist refraction lies on the same side as that of object. (A?B?) This image becomes the object for the concave mirror. For the mirror, \\ u = -\big(5 + \frac{20x}{x + 20}\big) = -\big(\frac{25x + 100}{x + 20}\big) \\ f = -10 cm \\ from mirror equation \\ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \frac{1}{v} = \frac{1}{-10} + \frac{x + 20}{25x + 100} \\ v = \frac{50(x + 4)}{3x - 20} \\ So, this image is formed towards left of the mirror. Again for second refraction in concave lens, \\ u = -\big[5 - \frac{50(x + 4)}{3x - 20}\big](assuming that image of mirror is formed between the lens and mirror)\\ v = +x (Since, the final image is produced on the object) \\ Using lens formula,\\ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \\ \frac{1}{x} + \frac{1}{5 - \frac{50(x + 4)}{3x - 20}} = \frac{1}{-20} \\ x = 60 cm\\ The object should be placed at a distance 60 cm from the lens further away from the mirror. So that the final image is formed on itself. 63 62. A converging lens of focal length 12 cm and a diverging mirror of focal length 7.5 cm are placed 5.0 cm apart with their principal axes coinciding. Where should an object be placed so that its image falls on itself? ##### Solution : It can be solved in a similar manner like question no.61, by using the sign conversions properly. Left as an exercise for the student. 64 63. A converging lens and a diverging mirror are placed at a separation of 15 cm. The focal length of the lens is 25 cm and that ofthe mirror is 40 cm. Where should a point source be placed between the lens and the mirror so that the light, after getting reflected by the mirror and then getting transmitted by the lens, comes out parallel to the principal axis ? ##### Solution : If the image in the mirror will form at the focus of the converging lens, then after transmission through the lens the rays of light will go parallel. Let the object is at a distance x cm from the mirror\\ \therefore u = -x cm;v = 25-15 = 10 cm(because focal length of the lens = 25 cm)\\ f = 40 cm; \frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$\\$ $\frac{1}{x} = \frac{1}{10} - \frac{1}{40}$$\\ x = \frac{400}{30} = \frac{40}{3}$$\\$ The object is at a distance $\big(15 - \frac{40}{3}\big) = \frac{5}{3}$ = 1.67 cm from the lense

65   64. A converging lens of focal length $15 cm$ and a converging mirror of focal length $10 cm$ are placed $50 cm$ apart with common principal axis. A point source is placed in between the lens and the mirror at a distance of $40 cm$ from the lens. Find the locations of the two images formed.

##### Solution : The object is placed in the focus of the converging mirror. There will be two images. a) One due to direct transmission of light through lens. b) One due to reflection and then transmission of the rays through lens. Case I :$(S^1)$ For the image by direct transmission,$\\$ u = -40 cm,f = 15 cm$\\$ $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$\\ \frac{1}{v} =\frac{1}{15} + \frac{1}{-40}$$\\$ v = 24 cm(left on lens)$\\$ Case II :$(S^")$ Since, the object is placed on the focus of mirror, after reflection the rays become parallel for the lens.$\\$ So, u = $\infty$$\\ f = 15 cm\\ \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$\\$ v = 15 cm(left of lens)$\\$

66   65. Consider the situation described in the previous problem. Where should a point source be placed on the principal axis so that the two images form at the same place ?

##### Solution : Let the source be placed at a distance ‘x’ from the lens as shown, so that images formed by both coincide.$\\$ for the lens, $\frac{1}{v} - \frac{1}{u} = \frac{1}{15}$$\\ v_f = \frac{15x}{x - 15}$$\\$ for the mirror u = -(50 - x), f = -10 cm$\\$ So $\frac{1}{v_m} + \frac{1}{-(50 - x)} = -\frac{1}{10}$$\\ so v_m = 50\\ \frac{15x}{x - 15} - \frac{10(50 - x)}{x - 40} = 50$$\\$ x = 30 cm$\\$ So, the source should be placed 30 cm from the lens.

67   66. A converging lens of focal length $15 cm$ and a converging mirror of focal length $10 cm$ are placed $50 cm$ apart. If a pin of length $2.0 cm$ is placed $30 cm$ from the lens farther away from the mirror, where will the final image form and what will be the size of the final image ?

##### Solution : given that $f_1 = 15 cm,f_m = 10 cm,h_0 = 2cm$$\\ The object placed 30 cm from lens \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$\\$ v =$\frac{uf}{u + f}$ $\\$ since u = -30 cm and f = 15 cm$\\$ so v = 30 cm$\\$ So, real and inverted image (A?B?) will be formed at 30 cm from the lens and it will be of same size as the object. Now, this real image is at a distance 20 cm from the concave mirror. Since, fm = 10 cm, this real image is at the centre of curvature of the mirror. So, the mirror will form an inverted image $A^1B^1$ at the same place of same size. Again, due to refraction in the lens the final image will be formed at AB and will be of same size as that of object. $A^1B^1$

68   67. A point object is placed on the principal axis of a convex lens $(f = 15 cm)$ at a distance of $30 cm$ from it. A glass plate $(t = 1.50)$ of thickness $1 cm$ is placed on the other side of the lens perpendicular to the axis. Locate the image of the point object.

##### Solution : for the lens f = 15 cm u = -30 cm $\\$ from the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$\\ \frac{1}{v} =\frac{1}{15} - \frac{1}{30}$$\\$ v = 30 cm$\\$ The image is formed at 30 cm of right side due to lens only. Again, shift due to glass slab is,$\\$ $\Delta t = \big(1 - \frac{1}{15}\big)1 [since \mu_g = 1.5 and t = 1 cm]$$\\ = 1 - (\frac{2}{3}) = 0.33 cm\\ then The image will be formed at 30 + 0.33 = 30.33 cm from the lens on right side. 69 68. A convex lens of focal length 20 cm and a concave lens of focal length 10 cm are placed 10 cm apart with their principal axes coinciding. A beam of light travelling parallel to the principal axis and having a beam diameter 5.0 mm, is incident on the combination. Show that the emergent beam is parallel to the incident one. Find the beam diameter of the emergent beam. ##### Solution : Let, the parallel beam is first incident on convex lens.\\ d = diameter of the beam = 5 mm\\ Now, the image due to the convex lens should be formed on its focus (point B)\\ So, for the concave lens,\\ u = +10 cm (since, the virtual object is on the right of concave lens)\\ f =-10 cm\\ So, \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$\\$ $\frac{1}{v} = \frac{1}{-10} + \frac{1}{10} = 0$$\\ v = \infty$$\\$ So, the emergent beam becomes parallel after refraction in concave lens. As shown from the triangles XYB and PQB,$\\$ $\frac{PQ}{XY} = \frac{RB}{ZB} =\frac{10}{20} = \frac{1}{2}$$\\ so the PQ = \frac{1}{2}\times5 = 2.5 mm\\ So, the beam diameter becomes 2.5 mm. Similarly, it can be proved that if the light is incident of the concave side, the beam diameter will be 1cm. 70 69. A diverging lens of focal length 20 cm and a converging lens of focal length 30 cm are placed 15 cm apart with their principal axes coinciding. Where should an object be placed on the principal axis so that its image is formed at infinity ? ##### Solution : Given that, f1 = focal length of converging lens = 30 cm\\ f2 = focal length of diverging lens = –20 cm\\ and d = distance between them = 15 cm\\ Let, F = equivalent focal length \\ so \frac{1}{F} =\frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1f2}$$\\$ $\frac{1}{30} + \big(-\frac{1}{20} \big) - \big(\frac{15}{30 - (200)} \big) = \frac {1}{120}$$\\ F = 120 cm\\ The equivalent lens is a converging one.\\ Distance from diverging lens so that emergent beam is parallel (image at infinity),\\ d_1 = \frac{dF}{f_1} = \frac{15\times120}{30} = 60 cm$$\\$ It should be placed 60 cm left to diverging lens$\\$ Object should be placed (120 – 60) = 60 cm from diverging lens.$\\$ Similarly $d_2 = \frac{dF}{f_2} = \frac{15\times120}{30} = 90 cm$$\\ So, it should be placed 90 cm right to converging lens.\\ Object should be placed (120 + 90) = 210 cm right to converging lens. 71 71. A point object is placed at a distance of 15 cm from a convex lens. The image is formed on the other side at a distance of 30 cm from the lens. When a concave lens is placed in contact with the convex lens, the image shifts away further by 30 cm. Calculate the focal lengths of the two lenses. ##### Solution : Let u = object distance from convex lens = –15 cm\\ v1 = image distance from convex lens when alone = 30 cm\\ f1 = focal length of convex lens\\ now \frac{1}{v} - \frac{1}{u} = \frac{1}{f_1}$$\\$ or $\frac{1}{f_1} = \frac{1}{30} - \frac{1}{-15} = \frac{1}{30} + \frac{1}{15}$$\\ or f_1 = 10 cm\\ Again, Let v = image (final) distance from concave lens = +(30 + 30) = 60 cm\\ v1 = object distance from concave lens = +30 m\\ f2 = focal length of concave lens\\ \frac{1}{v} - \frac{1}{v_1} = \frac{1}{f_1}$$\\$ $\frac{1}{f_1} = \frac{1}{60} - \frac{1}{30}$$\\ f_2 = -60 cm\\ So, the focal length of convex lens is 10 cm and that of concave lens is 60 cm. 72 70. A 5 mm high pin is placed at a distance of 15 cm from a convex lens of focal length 10 cm. A second lens of focal length 5 cm is placed 40 cm from the first lens and 55 cm from the pin. Find (a) the position of the final image, (b) its nature and (c) its size. ##### Solution : u = –15 cm, f = 10 cm\\ \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$\\$ $\frac{1}{v} - \big(-\frac{1}{15}\big) = -\frac{1}{10}$$\\ v = 30 cm\\ So, the final image is formed 10 cm right of second lens.\\ b) m for 1st lens :\frac{v}{u} = \frac{h_image}{h_object} = \frac{h_image}{5mm}$$\\$ $h_image$ = -10 mm (inverted)$\\$ Second lens : u = –(40 – 30) = –10 cm ; f = 5 cm$\\$ [since, the image of 1st lens becomes the object for the second lens].$\\$ $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$\\ \frac{1}{v} - \big(-\frac{1}{10}\big) = \frac{1}{5}$$\\$ m for second lens:$\\$ $\frac{v}{u} = \frac{h_image}{h_object} = \frac{10}{10}$$\\ So size of final image = 10 mm 73 74. A particle is moving at a constant speed V from a large distance towards a concave mirror of radius R along its principal axis. Find the speed of the image formed by the mirror as a function of the distance x of the particle from the mirror. ##### Solution : Given that, u = distance of the object = –x\\ f = focal length = \frac{–R}{2}$$\\$ and, V = velocity of object =$\frac{dx}{dt}$$\\ from mirror equation \frac{1}{-x}+\frac{1}{v} = -\frac{2}{R}$$\\$ $\frac{1}{v} = -\frac{2}{R} + \frac{1}{x} = \frac{R - 2x}{Rx}$$\\ v = \frac{Rx}{R-2x} = image distance\\ So velocity of the image is given by \\ v_1 = \frac{dv}{dt} = \frac{[\frac{d}{dt}(xR)(R - 2x)] - [\frac{d}{dt}(R - 2x)][xR]}{(R - 2x)^2}$$\\$ =$\frac{R[\frac{dx}{dt}(R - 2x)] - [-2\frac{dx}{dt}x]}{(R - 2x)^2} = \frac{R[v(R - 2x) + 2vx0}{(R - 2x)^2}$$\\ = \frac{VR^2}{2x - R^2} = \frac{R[VR - 2xV + 2xV}{(R - 2X)^2}$$\\$

74   72. Two convex lenses, each of focal length $10 cm$, are placed at a separation of $15 cm$ with their principal axes coinciding, (a) Show that a light beam coming parallel to the principal axis diverges as it comes out of the lens system. (b) Find the location of the virtual image formed by the lens system of an object placed far away. (c) Find the focal length of the equivalent lens. (Note that the sign of the focal length is positive although the lens system actually diverges a parallel beam incident on it).

##### Solution : a) The beam will diverge after coming out of the two convex lens system because, the image formed by the first lens lies within the focal length of the second lens.$\\$ b) for first conve lens $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$\\ \frac{1}{v} = \frac{1}{10} (since u = -\infty)$$\\$ or v =10 cm$\\$ for second convex lens $\frac{1}{v^1} = \frac{1}{10} + \frac{1}{-(15 - 10)} = \frac{-1}{10}$$\\ or v^1 = -10 cm\\ So, the virtual image will be at 5 cm from 1st convex lens.\\ c) If, F be the focal length of equivalent lens,\\ then \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1f_2}$$\\$ $\frac{1}{10} + \frac{1}{10} - \frac{15}{100} = \frac{1}{20}$$\\ F = 20 cm\\ 75 73. A ball is kept at a height h above the surface of a heavy transparent sphere made of a material of refractive index 1.t. The radius of the sphere is R. At t= 0, the ball is dropped to fall normally on the sphere. Find the speed of the image formed as a function of time for t < \sqrt{\frac{2h}{g}} • Consider only the image by a single refraction. ##### Solution : Let us assume that it has taken time ‘t’ from A to B.\\ AB = \frac{1}{2}gt^2$$\\$ BC= h - $\frac{1}{2}gt^2$$\\ This is the distance of the object from the lens at any time ‘t’.\\ here u = - ( h - \frac{1}{2}gt^2)\\ \mu_2 = \mu(given) and \mu_1 = i(air)$$\\$ so $\frac{\mu}{v} - \frac{1}{- (h - \frac{1}{2}gf^2)} = \frac{\mu - 1}{R}$ $\\$ $\frac{\mu}{v} = \frac{\mu - 1}{R} - \frac{1}{(h - \frac{1}{2}gt^2) }= \frac{(\mu - 1)( h - \frac{1}{2}gt^2) -R}{R( h - \frac{1}{2}gt^2) }$ $\\$ So, v = image distance at any time t = $\frac{\mu R(h - \frac{1}{2}gt^2) }{(\mu - 1)(h - \frac{1}{2}gt^2) - R}$ $\\$ So,velocity of the image = v = $\frac{dv}{dt} = \frac{d}{dt}\big[\frac{\mu R(h - \frac{1}{2}gt^2)}{(\mu - 1)(h - \frac{1}{2}gt^2) - R}\big]$$\\ = \frac{\mu R^2gt}{(\mu - 1)(h - \frac{1}{2}gt^2) - R} (can be find out) \\ 76 75. A small block of mass m and a concave mirror of radius R fitted with a stand lie on a smooth horizontal table with a separation d between them. The mirror together with its stand has a mass m. The block is pushed at t = 0 towards the mirror so that it starts moving towards the mirror at a constant speed V and collides with it. The collision is perfectly elastic. Find the velocity of the image (a) at a time t < \frac{d}{v}, (b) at a time t > \frac{d}{v}. ##### Solution : a) When t < \frac{d}{v}, the object is approaching the mirror\\ As derived in the previous question,\\ V_{image} = \frac{Velocity of object \times R^2}{[2 \times distance between them - R]^2} \\ V_{image} = \frac{V R^2}{[2(d - Vt) - R]^2} [at any time x = d - Vt] \\ b) After a time t > \frac{d}{V}, there will be a collision between the mirror and the mass. \\ As the collision is perfectly elastic, the object (mass) will come to rest and the mirror starts to move away with same velocity V. \\ At any time t > \frac{d}{V}, the distance of the mirror from the mass will be \\ x = V\big(t - \frac{d}{v}\big) = Vt - d \\ here u = -(Vt - d) = d - Vt; f = -\frac{R}{2} \\ so \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \frac{1}{v} = -\frac{1}{d - Vt } + \frac{1}{\frac{-R}{2}} = \big[\frac{R + 2(d -Vt)}{R(d -Vt)}\big] \\ v = -\big[\frac{R(d -Vt)}{R - 2(d - Vt)}\big] = image distance \\ so velocity of the image will be \\ V_{image} = \frac{d}{dt}(image distance) = \frac{d}{dt} \big[\frac{R(d - vt)}{R + 2(d - Vt)}\big] \\ Let y = (d - Vt) \\ \frac {dy} {dt} = -v \\ so V_{image} = \frac{d}{dt}\big[\frac{RY}{R + 2Y}\big] = \frac{(R + 2Y)(R(-V) - Ry(+2)(-V)}{R + 2y)^2} \\ = -Vr\big[\frac{R + 2y - 2y}{(R + 2y)^2}\big] = \frac{-V R^2}{(R + 2y)^2} \\ Since, the mirror itself moving with velocity V, \\ Absolute velocity of image = V = \big[ 1 - \frac{R^2}{(R + 2y)^2}\big] \\ = v\big[1 - \frac{R^2}{[2(Vt - d) - R^2\big]} \\ 77 76. A gun of mass M fires a bullet of mass m with a horizontal speed V. The gun is fitted with a concave mirror of focal length f facing towards the receding bullet. Find the speed of separation of the bullet and the image just after the gun was fired. ##### Solution : Recoil velocity of gun = v_g = \frac{mV}{M}$$\\$ at any time 't' position of the bullet w.r.t mirror = Vt + $\frac{mV}{M}t = \big(1 + \frac{m}{M}\big)vt$ $\\$ for the mirror u = -$\big(1 + \frac{m}{M}\big)vt = kVt$ $\\$ v = position of the image $\\$ From lens formula, $\\$ $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ $\\$ $\frac{1}{v} = \frac{1}{-f} + \frac{1}{kVt} = \frac{1}{kvt} - \frac{1}{f} = \frac{f - kVt}{kVtf}$ $\\$ let $\big(1 + \frac{m}{M} = k\big)$ $\\$ so v = $\frac{kVft}{-kVt + f} = \frac{kVft}{f - kVt}$ $\\$ So, velocity of the image with respect to mirror will be,$\\$ $v_1 =\frac{dv}{dt} = \frac{d}{dt}\big[\frac{kVtf}{f - kVt}\big] = \frac{(f - kVt)(kVf - kVtf(kv)}{(f-kVt)^2} = \frac{kVt^2}{(f - kVt)^2}$$\\ Since, the mirror itself is moving at a speed of mV/M and the object is moving at ‘V’, the velocity of separation between the image and object at any time ‘t’ will be,\\ v_s =V + \frac{mv}{M} + \frac{kvf^2}{(f - kVt)^2}$$\\$ when t = 0$\\$ $V_s = V + \frac{m V}{M} + kV $$\\ = V + \frac{m}{M}V + \big(1 + \frac{m}{M}\big)V$$\\$ = $2\big(1 + \frac{m}{v}\big)V$

78   77. A mass m = 50 g is dropped on a vertical spring of spring constant 500 N/m from a height $h = 10 cm$ as shown in figure $(18-E14)$. The mass sticks to the spring and executes simple harmonic oscillations after that. A concave mirror of focal length $12 cm$ facing the mass is fixed with its principal axis coinciding with the line of motion of the mass, its pole being at a distance of $30 cm$ from the free end of the spring. Find the length in which the image of the mass oscillates.

##### Solution : Due to weight of the body suppose the spring is compressed by which is the mean position of oscillation$\\$. $m = 50\times10^3 kg, g = 10 ms^{-2}, k = 500 Nm^{-2}, h = 10 cm = 0.1 m$ $\\$ For equilibrium, mg = kx then x = $\frac{mg}{k} = 10^{-3}m = 0.1 cm$ $\\$ So, the mean position is at 30 + 0.1 = 30.1 cm from P (mirror). $\\$ Suppose, maximum compression in spring is $\delta$ $\\$. Since, E.K.E. – I.K.E. = Work done $\\$ then 0 – 0 = mg(h + $\delta) – \frac{1}{2}k\delta^2$(work energy principle) $\\$ then mg(h + $\delta) = \frac{1}{2} k\delta^2$ $\\$ = $50\times10^-3 \times 10(0.1 + \delta) = \frac{1}{2}500\delta^2$ $\\$ So, $\delta= \frac{05\pm\sqrt{0.25 + 50}}{2\times250} = 0.015 m = 1.5 cm$ $\\$ From figure B,$\\$ Position of B is 30 + 1.5 = 31.5 cm from pole.$\\$ Amplitude of the vibration = 31.5 – 30.1 – 1.4.$\\$ Position A is 30.1 – 1.4 = 28.7 cm from pole.$\\$ For A u = –31.5, f = –12 cm$\\$ $\frac{1}{v} =\frac{1}{f} - \frac{1}{u} = -\frac{1}{12} + \frac{1}{31.5}$ $\\$ $v_A = -19.38 cm$$\\ For B f = -12 cm,u = -28.7cm\\ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -\frac{1}{12} + \frac{1}{28.7}$$\\$ $v_B = -20.62 cm$ $\\$ the image vibrates in length (20.62 - 19.38) = 1.24 cm$\\$

79   78) Two concave mirrors of equal radii of curvature R are fixed on a stand facing opposite directions. The whole system has a mass m and is kept on a frictionless horizontal table (figure 18-E15).$\\$ Two blocks A and B, each of mass m, are placed on the two sides of the stand. $At t = 0$, the separation between $A$ and the mirrors is $2 R$ and also the separation between $B$ and the mirrors is $2 R$. The block B moves towards the mirror at a speed v. All collisions which take place are elastic. Taking the original position of the mirrors-stand system to be $x = 0$ and X-axis along AB, find the position of the images of $A$ and $B$ at $t = \frac{R}{v}$$\\ b) \frac{3R}{v}$$\\$ c) $\frac{5R}{v}$

##### Solution : a) In time, t = R/V the mass B must have moved $(v\times\frac{R}{v}$) = R closer to the mirror stand$\\$ So, For the block B :$\\$ u = -R,f = $\frac{-R}{2}$$\\ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -\frac{2}{R} + \frac{1}{R} = -\frac{1}{R}$$\\$ v = -R at the same place$\\$ for the block A:u = -2R,f = $\frac{-R}{2}$$\\ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{-2}{R} + \frac{1}{2R} = \frac{-3}{2R}$$\\$ v = $\frac{-2R}{3}$ image A at $\frac{2R}{3}$ from PQ in the x radiation$\\$ So, with respect to the given coordinate system,$\\$ Position of A and B are $\frac{-2R}{3}$, R respectively from origin.$\\$ b) When $t = 3\frac{R}{v}$, the block B after colliding with mirror stand must have come to rest (elastic collision) and the mirror have travelled a distance R towards left form its initial position.$\\$ So, at this point of time,$\\$ For block A :$\\$ u =-R,f = $\frac{-R}{2}$$\\ Using lens formula, v = –R (from the mirror), So, position xA = –2R (from origin of coordinate system) For block B : Image is at the same place as it is R distance from mirror. Hence, position of image is ‘0’. Distance from PQ (coordinate system)\\ positions of images of A and B are = –2R, 0 from origin.\\ c) Similarly, it can be proved that at time t = 5\frac{R}{v},\\ the position of the blocks will be –3R and –4\frac{R}{3} respectively. 80 79. Consider the situation shown in figure (18-E16). The elevator is going up with an acceleration of 2'00 (\frac{m}{s})^ 2 and the focal length of the mirror is 12'0 cm. All the surfaces are smooth and the pulley is light. The mass-pulley system is released from rest (with respect to the elevator) at t = 0 when the distance of B from the mirror is 42'0 cm. Find the distance between the image of the block B and the mirror at t = 0'200 s. Take g= 10 (\frac{m}{s})^2.\\ ##### Solution : Let a = acceleration of the masses A and B (w.r.t. elevator). From the freebody diagrams,\\ T - mg + ma - 2m = 0;.........(1)\\ similarly T - ma = 0.......(2)\\ from (1) and (2) 2ma - mg - 2m = 0;\\ 2ma = m(g + 2)\\ a = \frac{10 + 2}{2} = \frac{12}{2} = 6\frac{m}{s^-2}$$\\$ so distance from mirror, u = -(42 - 120 = -30 cm; f = 12 cm$\\$ from mirror equation $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$\\ \frac{1}{v} + \big(-\frac{1}{30}\big) = \frac{1}{12}$$\\$ v = 8.57 cm$\\$ Distance between image of block B and mirror = 8.57 cm.

81   15. A pole of length $1'00 m$ stands half dipped in a swimming pool with water level $50'0 cm$ higher than the bed. The refractive index of water is 1'33 and sunlight is coming at an angle of 45° with the vertical. Find the length of the shadow of the pole on the bed.