Ncert Solutions Physics Class 12th

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^$°$

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

Focal length of the concave lens, $f_2$ = -20 cm

Distance between two lenses, d = 8.0 cm

When the parallel beam of light is incident on the convex lens first:

According to lens formula, we have:

$\frac{1}{v_1}$ - $\frac{1}{u_1}$ = $\frac{1}{f_1}$ , where u = object distance = $\infty$ and $v_1$ = Image distance

$\frac{1}{v_1}=$ $\frac{1}{f_1}+\frac{1}{u_1}$ = $\frac{1}{30}$ + $\frac{1}{\infty }$

$v_1=30cm$

The image will act as a virtual object for the concave lens. Applying lens formula to the concave lens, we have:

$\frac{1}{v_2}$ - $\frac{1}{u_2}$ = $\frac{1}{f_2}$ ,

where $u_2$ = object distance = $v_1-d$ = 30 – 8 = 22 cm and

$v_2$ = image distance

$\frac{1}{v_2}=\frac{1}{f_2}+$ $\frac{1}{u_2}$ = $\frac{1}{-20}$ + $\frac{1}{22}$

$v_2=$ - 220 cm

The parallel incident beam appears to diverge from a point that is 220 - $\frac{d}{2}$ = 216 cm from the centre of the combination of two lenses.

When the parallel beam of light is incident, from the left on the concave lens first:

According to the lens formula,

$\frac{1}{v_2}$ - $\frac{1}{u_2}$ = $\frac{1}{f_2}$

$\frac{1}{v_2}=\frac{1}{f_2}+\frac{1}{u_2}$ , where $u_2$ = Object distance = - $\infty$

$\frac{1}{v_2}=\frac{1}{*20}+\frac{1}{-\mathrm{}\infty }$

$v_2=$ - 20 cm

The image will act as a real object for the convex lens.

Applying lens formula to the convex lens, we have

$\frac{1}{v_1}$ - $\frac{1}{u_1}$ = $\frac{1}{f_1}$ , where $u_1$ = object distance = - (20 + d) = - 28 cm and $v_1$ = Image distance

$\frac{1}{v_1}=\frac{1}{f_1}$ + $\frac{1}{u_1}$ = $\frac{1}{30}$ + $\frac{1}{-28}$

$v_1$ = - 420 cm

Hence, the parallel incident beam appear to diverge from a point that is 420 – 4 = 416 cm from the left of the centre of the combination of the two lenses.

The answer does not depend on the side of the combination at which the parallel beam of light is incident. The notion of effective focal length does not seem to be useful for this combination.

Height of the image, $h_1$ = 1.5 cm

Object distance from the side of the convex lens, $u_1$ = - 40 cm

$\left|u_1\right|$ = 40 cm

According to lens formula:

$\frac{1}{v_1}$ - $\frac{1}{u_1}$ = $\frac{1}{f_1}$ , where $v_1$ = Image distance

$\frac{1}{v_1}$ = $\frac{1}{f_1}+\frac{1}{u_1}$ , where $v_1$ is the image distance

$\frac{1}{v_1}$ = $\frac{1}{30}+\frac{1}{-40}$

$v_1$ = 120 cm

$\mathrm{M}\mathrm{a}\mathrm{g}\mathrm{n}\mathrm{i}\mathrm{f}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n},m_1=$ $\frac{v_1}{\left|u_1\right|}$ = 3

Hence the magnification due to the convex lens is 3.

The image formed by the convex lens acts as an object for the concave lens.

According to lens formula,

$\frac{1}{v_2}$ - $\frac{1}{u_2}$ = $\frac{1}{f_2}$ , where $u_2$ is the object distance = + (120 – 8) = + 112 cm

$\frac{1}{v_2}=$ $\frac{1}{f_2}+\frac{1}{u_2}$ = $\frac{1}{-20}$ + $\frac{1}{112}$

$v_2$ = - 24.35 cm

Magnification, $m_2$ = $\left|\frac{v_2}{112}\right|$ = 0.217

The magnification of combination of two lenses given as m = $m_1*m_2$ = 0.652

The magnification of the combination is also expressed as

$\frac{h_2}{h_1}$ = 0.652, $h_2$ = 0.652 $*h_1$ = 0.652 $*1.5$ = 0.978 cm

Hence, the height of the image is 0.978 cm.

9.18 (a) Yes, Plane and convex mirrors can produce real images as well, if the object is virtual.

i.e. if the light rays converging at a point behind a plane mirror or a convex mirror, they are reflected to a point on a screen placed in front of the mirror, then a real image will be formed.

(b) No. A virtual image is formed when light rays diverge. The convex lens of the eye causes these divergent rays to converge at the retina. In this case, the virtual image serves as an object for the lens to produce a real image.

(c) Water being the denser medium than air, the light rays will deviate and the fisherman will appear to be taller to the diver.

(d) The apparent depth of the tank water changes when viewed obliquely. This is because light bends on travelling from one medium to another. The apparent depth of the tank when viewed obliquely is less than the near normal viewing.

(e) No, diamond is used as cutter because of its hardness, not refractive index.

9.17 Refractive index of the glass fibre, ${\mu }_1$ = 1.68

Refractive index of outer covering of the pipe, ${\mu }_2$ = 1.44

Given:-

Angle of incidence = i, Angle of refraction = r, Angle of incidence at the interface = i'

The refractive index of the inner core-outer core interface, $\mu$ is given as

$\mu =\frac{{\mu }_2}{{\mu }_1}$ = $\frac{1}{\sin ?i\text{'}}$

$\frac{1}{\sin ?i\text{'}}=\frac{1.44}{1.68}$ or $\sin ?i\text{'}$ = $\frac{1.44}{1.68}$

i' = 59 $°$

For the critical angle, total reflection take place only when i > i' i.e. i > 59 $°$

Maximum angle of reflection, ${\tau }_{max}$ = 90 $°-i\text{'}$ = 31 $°$

Let $i_{max}$ be the maximum angle of incidence and $r_{max}$ be the maximum angle of reflection.

${\mu }_1=\frac{\sin ?i_{max}}{\sin ?r_{max}}$ or $\sin ?i_{max}={\mu }_1*\sin ?r_{max}$

$\sin ?i_{max}=$ 1.68 $*sin31°$

$i_{max}=59.91°$ $\approx 60°$

The entire rays incident at angles lying in the range of 0 < i < 60 will suffer total internal reflection.

If the outer covering is absent, then:

Refractive index of the outer pipe, ${\mu }_1$ = Refractive index of air = 1

For the angle of incidence, i = 90 $°$ , we can write Snell's law at 'air – pipe' interface as :

$\frac{sini}{\sin ?r}$ = ${\mu }_2$ = 1.68

$\sin ?r=$ $\frac{\sin ?i}{1.68}$ = $\frac{\sin ?90°}{1.68}$ = $\frac{1}{1.68}$

r = 36.53 $°$

i' = 90 $°-$ 36.53 $°$ = 53.47 $°$

Since i' > r, all incident rays will suffer total internal reflection.