# Concept Of Physics Geometrical Optics

#### 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.

$u$ = -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.

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.

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

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} $$\\ then A^{1}b^{1 }= \frac{AB } {200 } = \frac{ 16} { 200} = 0.08 cm = 0.8 m. 6. 6. A 3 cm tall object is placed at a distance of 7.5 cm from a convex mirror of focal length 6 cm. Find the location, size and nature of the image. 7. 7. A U-shaped wire is placed before a concave mirror having radius of curvature 20 cm as shown in figure (18-E1). Find the total length of the image. Answer 7 None Suppose the height is h. \\ At earth station F = \frac{ GMm} { R^{2} }$$\\$ M = mass of earth.$\\$ m = mass of satelite$\\$ R = radius of earth$\\$ f = $\frac{ GMm} { 2R^{2} }$$\\ then 2R^{2} = (R + h)^2$$\Rightarrow$ $R^{2}$ - $h^{2}$ - 2Rh = 0$\\$ then $h^{2} + 2Rh-R^{2}$ = 0$\\$ $H=\frac{\big(-2R\pm\sqrt{4R^2+4R^2}\big)}{2}$ = $\frac{-2R\pm2\sqrt{2R}}{2}$ $\\$ =$-R\pm\sqrt{2R}$ = $R{(\sqrt2 - 1)}$$\\$ = 6400 x (0.414)$\\$ = 2649.6 = 2650 km

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.

Two charged particle placed at a sehortion 2m. exert a force of 20 m.$\\$ $F_1$ = 20 N.$F_2$= ? r1 = 20 cm.r2 = 25 cm. $\\$ since F = $\frac{ 1} {4\pi\varepsilon_0}$ $\frac{ q_1q_2} { r^2 }$, F = $\frac{1 } { r^2 }$ $\\$ $\frac{f_1 } { f_2 }$ = $\frac{ r_2^2} { r_1^2}$ $\\$ $F_2$ =$F_1$ x $\big(\frac{r_1}{r_2}\big)^2$ = 20 x $\big(\frac{20}{25}\big)^2$ = $20 \times \frac{16}{25}$ = 12.8 N = 13 N.

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 $10 5 km4$.