Physics

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

9.26 Though the image size is bigger than the object, the angular size of the image is equal to the angular size of the object. A magnifying glass helps one see the object placed closer than the least distance of distinct vision (i.e. 25 cm). A closer object causes a larger angular size. A magnifying glass provides angular magnification. Without magnification, the object cannot be placed closer to the eye. With magnification, the object can be placed much closer to the eye.

Yes, the angular magnification changes. When the distance between the eye and a magnifying glass is increased, the angular magnification decreases a little. This is because the angle subtended at the eye is lightly less than the angle subtended at the lens. Image distance does not have any effect on angle magnification.

The focal length of a convex lens cannot be decreased by a greater amount. This is because making lenses having very small focal length is not easy. Spherical and chromatic aberrations are produced by a convex lens having a very small focal length.

The angular magnification produced by the eyepiece of a compound microscope is [( $\frac{25}{f_e}$ ) + 1], where $f_e$ = focal length of the eyepiece

It can be inferred that if $f_e$ is small, then angular magnification of the eyepiece will be large.

The angular magnification of the objective lens of a compound microscope is given as $\frac{1}{(\left|u_0\right|f_{0)}}$ , where $u_0$ = Object distance for the objective lens and $f_0$ = Focal length of the objective.

The magnification is large when $u_0$ > $f_0$ . In the case of a microscope, the object is kept close to the objective lens. Hence, the object distance is very little. Since $u_0$ is small, $f_0$ will be even smaller. Therefore, $f_e$ and $f_0$ are both small in the given condition.

When we place our eyes too close to the eyepiece of a compound microscope, we are unable to collect much refracted light. As a result, the field of view decreases substantially. Hence, the clarity of the image gets blurred.

The best position of the eye for viewing through a compound microscope is at the eye-ring attached to the eyepiece. The precise location of the eye depends on the separation between the objective lens and the eyepiece.

9.23 Area of each square, A = 1 ${mm}^2$

Object distance, u = - 9 cm

Focal length of the converging lens, f = 10 cm

For image distance v, the lens formula can be written as

$\frac{1}{f}$ = $\frac{1}{v}$ - $\frac{1}{u}$

$\frac{1}{10}$ = $\frac{1}{v}$ - $\frac{1}{-9}$

$\frac{1}{v}$ = $\frac{1}{10}-\frac{1}{9}$

v = -90 cm

Magnification, m = $\frac{v}{u}$ = $\frac{-90}{-9}$ = 10

Therefore the area of each square of the virtual image

= 10 $*10{mm}^2$ = 100 ${mm}^2=1{cm}^2$

Magnifying power of the lens = $\frac{d}{\left|u\right|}$ = $\frac{25}{9}$ = 2.8

The magnification in (a) is not the same as the magnifying power in (b). The magnification magnitude is = $\frac{v}{u}$ and the magnifying power is = $\frac{d}{\left|u\right|}$ . The two quantities will be equal when the image is formed at the near point (25 cm)

9.22 The incident, refracted and emergent rays associated with a glass prism ABC is shown in the adjoined figure.

Angle of the prism, $\angle A$ = 60 $°$

Refractive index of the prism, $\mu$ = 1.524

Let $\angle i_1$ be the incident angle, ${\angle r}_1$ be the refracted angle, $\angle r_2$ be the incidence angle on face AC and $\angle e$ be the emergent angle from the prism = 90 $°$

According to Snell's law, for face AC, we can have:

$\frac{\sin ?e}{\sin ?r_2}$ = $\mu$

$\sin ?r_2=\frac{\sin ?e}{\mu }$ = $\frac{\sin ?90°}{1.524}$ = 0.656

${\angle r}_2$ = 41 $°$

From the Δ ABC, $\angle A=$ $\angle r_1$ + ${\angle r}_2$ . Hence ${\angle r}_1$ = 60 $°-$ 41 $°=19°$

According to Snell's law, for face AB, we have

$\mu =\frac{sin{i}_1}{\sin ?r_1}$ or $\sin ?i_1=\mu *\sin ?r_1$ = 1.524 $*\sin ?19°=0.496$

$\angle i_1$ = 29.75 $°$

Hence the angle of incidence is 29.75^$°$