Ncert Solutions Physics Class 12th

9.32 Focal length of the convex lens, $f_1$ = 30 cm

The liquid acts as a mirror, focal length of the liquid = $f_2$

Focal length of the system (convex lens + liquid), $f$ = 45 cm

For a pair of optical systems placed in contact, the equivalent focal length is given as

$\frac{1}{f}$ = $\frac{1}{f_1}$ + $\frac{1}{f_2}$ or $\frac{1}{f_2}=\frac{1}{f}-\frac{1}{f_1}$ = $\frac{1}{45}$ - $\frac{1}{30}$

$f_2=$ - 90 cm

Let the refractive index of the lens be ${\mu }_1$ and the radius of curvature of one surface be R

Hence, the radius of curvature of the other surface is –R

R can be obtained by using the relation

$\frac{1}{f_1}$ = ( ${\mu }_1-1\left)\right(\frac{1}{R}$ + $\frac{1}{\left|R\right|}$ ) = (1.5 – 1)( $\frac{2}{R})$

$\frac{1}{30}$ = $\frac{1}{R}$ , so R = 30 cm

Let the refractive index of the liquid be ${\mu }_2$

The radius of curvature of the liquid on the side of the plane mirror = $\infty$

Radius of curvature of the liquid on the side of the lens, R = -30 cm

The value of ${\mu }_2$ can be calculated using the relation,

$\frac{1}{f_2}$ = ( ${\mu }_2-1\left)\right(\frac{1}{-R}$ - $\frac{1}{\infty }$ )

$\frac{-1}{90}$ = ( ${\mu }_2-1\left)\right(\frac{1}{30}$ - 0)

${\mu }_2$ - 1 = $\frac{1}{3}$

${\mu }_2=\frac{4}{3}$ = 1.33

Hence, the refractive index of the liquid is 1.33.

9.30 Distance between the objective mirror and the secondary mirror, d = 20 mm

Radius of curvature of objective mirror,  $R_1$ = 220 mm

Hence focal length of the objective mirror,  $f_1$ = $\frac{R_1}{2}$ = 110 mm

Radius of curvature of secondary mirror,  $R_2$ = 140 mm

Hence focal length of the objective mirror,  $f_2$ = $\frac{R_2}{2}$ = 70 mm

The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for the secondary mirror. Hence, the virtual object distance for the secondary mirror,

u = $f_1-d$ = 110 – 20 = 90 mm

Applying the mirror formula for the secondary mirror, we can calculate the image distance ( $v$ ) as:

$\frac{1}{v}$ + $\frac{1}{u}$ = $\frac{1}{f_2}$ or $\frac{1}{v}$ = $\frac{1}{f_2}-\frac{1}{u}$ = $\frac{1}{70}$ - $\frac{1}{90}$

$v$ = 315 mm

Hence, the final image will be formed 315 mm away from the secondary mirror.

9.29 Focal length of the objective lens, $f_o$ = 140 cm

Focal length of the eyepiece, $f_e$ = 5 cm

Least distance of distinct vision, d = 25 cm

In normal adjustment, the separation between the objective lens and the eyepiece

= $f_o+f_e=140+5=145cm$

Height of the tower $h_1=100m$

Distance of the tower (object) from the telescope, u = 3 km = 3000 m

The angle subtended by the tower at the telescope is given as : $\theta =\frac{h}{u}$ = $\frac{100}{3000}$ = $\frac{1}{30}$ rad

The angle subtended by the image produced by the objective lens is given as $\theta =\frac{h_2}{f_o}$ , where $h_2$ = height of the image of the tower formed by the objective lens

So, $h_2$ = $\theta *f_o$ = $\frac{1}{30}*140$ =4.7 cm

Therefore, the objective lens forms a 4.7 cm tall image of the tower.

Image is formed at a distance, d = 25 cm

The magnification of the eyepiece is given by the relation

m = 1 + $\frac{d}{f_e}$ = 1 + $\frac{25}{5}$ = 6

Height of the final image = m $*h_2$ = 6 $*4.7=28.2cm$

9.27 Focal length of the objective lens, $f_0$ = 1.25 cm

Focal length of the eyepiece, $f_e$ = 5 cm

Least distance of distinct vision, d = 25 cm

Angular magnification of the compound microscope = 30X

Total magnifying power of the compound microscope, m = 30

The angular magnification of the eyepiece is given by the relation:

$m_e$ = (1 + $\frac{d}{f_e}$ ) = (1 + $\frac{25}{5})$ = 6

The angular magnification of the objective lens ( $m_o$ ) is related to $m_e$ by the equation

m = $m_o{m}_e$ or

$m_o$ = $\frac{m}{m_e}$ = $\frac{30}{6}$ = 5

We also have relation

$m_o=\frac{Imagedistanceforobjectivelens\left(v_o\right)}{Objectdistanceforobjectivelens(-u_o)}$

5 = $\frac{v_o}{-u_o}$ or $v_o=-5u_o$ ……….(1)

Applying lens formula for the objective lens

$\frac{1}{f_0}$ = $\frac{1}{v_o}$ - $\frac{1}{u_o}$ or $\frac{1}{1.25}$ = $\frac{1}{-5u_o\mathrm{}}$ - $\frac{1}{u_o}$

$\frac{1}{1.25}=\frac{-6}{5u_o}$

$u_o=$ $-\frac{6*1.25}{5}$ = $-1.5cm$

From equation (1), $v_o=-5u_o$ = 7.5 cm

The object should be placed 1.5 cm away from the objective lens to obtain desired magnification.

Applying the lens formula for the eyepiece,

$\frac{1}{v_e}-\frac{1}{u_e}$ = $\frac{1}{f_e}$ , where $v_e$ = Image distance for eyepiece = -d = -25 cm,

$u_e$ = Object distance for the eyepiece

$\frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}$ = $\frac{1}{-25}$ - $\frac{1}{5}$

$u_e=-\frac{25}{6}$ = $-4.17cm$

Separation between the objective lens and the eyepiece = $\left|u_e\right|$ + $\left|v_o\right|$ = 4.17 + 7.5 = 11.67 cm