Concept Of Physics Geometrical Optics

H C Verma

Concept Of Physics

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 $10 5 km4$.