NCERT Solutions Class 12 Physics Chapter 9: Ray Optics and Optical Instruments

NCERT Solutions for Class 12 Physics Ch 9 Ray Optics consists of step-by-step explanations for all textbook questions, which maks it easier for students to comprehend difficult concepts and help develop problem-solving techniques. 

Class 12 Physics Ch 9 Ray Optics NCERT Solutions also help students clarify basic principles to improve conceptual clarity, and help students revise properly before exams. Since competitive exams like JEE and NEET include questions related to the Class 12 Physics NCERT Topics and concepts, these NCERT solutions PDFs are an excellent resource to thoroughly prepare for the exam.

Class 12 Physics Ch 9 Ray Optics helps students understand the fundamental concepts on how light interacts with optical systems such as mirrors and lenses. This chapter helps students to build the basis for many real-world applications, such as vision correction, telescopes, microscopy, and fiber optics. Also, there are important concepts like refraction, reflection, and total internal reflection are often asked in board exams and competitive exams like JEE and NEET, so being aware of this Class 12 Physics Chapter 9 Ray Optics is important for scoring well.

Students can access the NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics PDF from different educational platforms like Shiksha and other online resources. 

To be able to donwload the NCERT Solutions PDF for Class 12 Physics Chapter 9 Ray Optics candidates must visit the homepage of Shiksha's Class 12 Physics Chapter 9 Ray Optics and click on the “NCERT Physics Class 12th Solution PDF - Ray Optics Chapter Download” to access the PDF file. 

These NCERT Solution PDFs help students to study offline, to be able to understand important problems, and practice numerical questions without needing an internet connection at any time. This help in revision and imporve their understanding of the crucial topics and concepts.

The are various important topics often asked in the competitve exams an even in board exams related to the Class 12 Physics Ch 9 Ray Optics such as:

• Reflection of Light

• Refraction of Light

• Total Internal Reflection

• Lens Maker's Formula

• Power of a Lens

• Combination of Lenses

• Optical Instruments

• Huygens' Principle

These topics are important to understand the behavior of light and its applications in optical devices. Understand these concepts will help students answer questions related to theory and numerical problems in board and entrance exams.

9.15 For a concave mirror, the focal length, f < 0

When the object is placed on the left side of the mirror, the object distance, u is negative

For image distance v, we can write the lens formula as:

$\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ = $\frac{1}{f}-\frac{1}{u}$ (since u is negative) ……….(1)

The object lies between f and 2f, i.e. 2f < u < f

or, $\frac{1}{2f}>\frac{1}{u}>\frac{1}{f}$

or, $-\frac{1}{2f}<-\frac{1}{u}<-\frac{1}{f}$

or, $\frac{1}{f}$ $-\frac{1}{2f}<\frac{1}{f}-\frac{1}{u}$ < 0 …………(2)

Using equation (1), we get

$\frac{1}{2f}<\frac{1}{v}$ < 0

Therefore, $\frac{1}{v}$ is negative, hence v is negative

$\frac{1}{2f}<\frac{1}{v}or$ 2f > v or –v > -2f . Hence, the image lies beyond 2f

For a convex mirror, the focal length, the focal length f > 0

When the object is placed on the left side of the mirror, the object distance (u) is negative (< 0)

For image distance v, we have the mirror formula:

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

Using equation (2), we can conclude that:

$\frac{1}{v}<0$ or v > 0

Thus, the image is formed on the back side of the mirror.

Hence, a convex mirror always produces a virtual image, regardless of the object distance.

For a convex mirror, the focal length, the focal length f > 0

When the object is placed on the left side of the mirror, the object distance (u) is negative (< 0)

For image distance v, we have the mirror formula:

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

But we have u < 0, then

$\frac{1}{v}>\frac{1}{f}$ or v < f

$\mathrm{H}$ ence, the image formed is diminished and it is located between the focus (f) and the pole.

For a concave mirror, the focal length (f) is negative, f < 0

When the object is placed on the left side of the mirror, the object distance (u) is negative, u < 0

It is placed between the focus (f) and the pole, so f > u > 0

So, $\frac{1}{f}$ < $\frac{1}{u}$ < 0 or $\frac{1}{f}$ - $\frac{1}{u}$ < 0

For image distance v, we have the formula:

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

$\frac{1}{v}$ < 0 or v > 0

The image is formed on the right side of the mirror, hence it is a virtual image.

For u < 0 and v > 0, we can write $\frac{1}{u}$ > $\frac{1}{v}$ or v > u

Magnification, m = $\frac{v}{u}$ > 1

Hence, the formed image is enlarged.

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 $\times 10{mm}^2$ = 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)

9.6 The angle of minimum deviation, ${\delta }_m$ = 40 $°$

Angle of prism, A = 60 $°$

Let the refractive index of water, $=1.33$ , and the refractive index of prism material = $\mu \text{'}$

The angle of deviation is related to refractive index $\mu \text{'}$ is given as

${\mu }^{\text{'}}=\frac{\sin ?\frac{(A+{\delta }_m)}{2}}{\sin ?\frac{A}{2}}$ = $\frac{\sin ?\frac{(60°+40°)}{2}}{\sin ?\frac{60°}{2}}$ = $\frac{\sin ?50°}{\sin ?30°}$ = 1.532

So the refractive index of prism material is 1.532

Since the prism is placed in water, let ${\delta }_m^{\text{'}}$ be the new angle of minimum deviation.

The refractive index of glass with respect to water is given by the relation:

${\mu }_g^w$ = $\frac{{\mu }^{\text{'}}}{\mu }$ = $\frac{\sin ?\frac{(A+{\delta }_m^{\text{'}}\mathrm{})}{2}}{\sin ?\frac{A}{2}}$

$\frac{1.532}{1.33}$ = $\frac{\sin ?\frac{(60°+{\delta }_m^{\text{'}}\mathrm{})}{2}}{\sin ?\frac{60°}{2}}$

1.152 $\times \sin ?30°$ = $\sin ?\frac{(60°+{\delta }_m^{\text{'}}\mathrm{})}{2}$

$\frac{(60°+{\delta }_m^{\text{'}}\mathrm{})}{2}={\sin }^{-1}?0.576$ = 35.2 $°$

${\delta }_m^{\text{'}}=2\times$ 35.2 $°$ - 60 $°$ = 10.33 $°$

Hence the new minimum angle of deviation is 10.33$°$

9.13 Focal length of the objective lens,  $f_o$ = 144 cm

Focal length of the eyepiece,  $f_e$ = 6.0 cm

The magnifying power of the telescope, m = $\frac{f_o}{f_e}$ = $\frac{144}{6}$ = 24

The separation between the eyepiece and objective lens = $f_e+f_o$ = 6 + 144 = 150 cm

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 = $m_1\times m_2$ = 0.652

The magnification of the combination is also expressed as

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

Hence, the height of the image is 0.978 cm.

9.14 Focal length of the objective lens, $f_o$ = 15 m = 1500 cm

Focal length of the eyepiece, $f_e$ = 1.0 cm

The angular magnification of the telescope is given by,

m = $\frac{f_o}{f_e}$ = $\frac{1500}{1}$ = 1500

Hence the angular magnification of the telescope is 1500

Diameter of the moon, $d_m$ = 3.48 $\times {10}^6$ m

Radius of the lunar orbit, $r_l$ = 3.8 $\times {10}^8$ m

Let $d_1$ be the diameter of the image of the moon formed by the objective lens.

The angle subtended by the diameter of the moon is equal to the angle subtended by the image. Hence,

$\frac{d_m}{r_l}$ = $\frac{d_1\mathrm{}}{f_o\mathrm{}}$

$d_1=\frac{d_m\times f_o}{\mathrm{}{r}_l}$ = $\frac{3.48\mathrm{}\times {10}^6\times 1500\mathrm{}}{3.8\mathrm{}\times {10}^8}$ = 13.74 cm

Hence the diameter of the moon’s image is 13.74 cm.

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 $\sin ?i_{max}={\mu }_1\times \sin ?r_{max}$

$\sin ?i_{max}=$ 1.68 $\times sin31°$

$i_{max}=59.91°$ $\approx 60°$

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.

9.11 Focal length of the objective lens, $f_1$ = 2.0 cm

Focal length of the eyepiece, $f_2$ = 6.25 cm

Distance between the objective lens and the eyepiece, d = 15 cm

Least distance of distinct vision, d’ = 25 cm

Hence, image distance for the eyepiece, $v_2$ = - 25 cm

Let the object distance for the eyepiece be = $u_2$

According to lens formula, we get

$\frac{1}{v_2}$ - $\frac{1}{u_2}$ = $\frac{1}{f_2}$ or $\frac{1}{u_2}=\frac{1}{v_2}-\frac{1}{f_2}$ or $\frac{1}{u_2}=\frac{1}{-25}-\frac{1}{6.25}$

$u_2=-5cm$

Image distance for the objective lens, $v_1$ = d + $u_2$ = 15 – 5 = 10 cm

Let the object distance for the eyepiece be = $u_1$

According to lens formula, we get

$\frac{1}{v_1}$ - $\frac{1}{u_1}$ = $\frac{1}{f_1}$ or $\frac{1}{u_1}=\frac{1}{v_1}-\frac{1}{f_1}$ or $\frac{1}{u_1}=\frac{1}{10}-\frac{1}{2}$

$u_1=-2.5cm$

Magnitude for the object distance, $\left|u_1\right|$ = 2.5 cm

The magnifying power of a compound microscope is given by the relation,

m = $\frac{v_1}{\left|u_1\right|}$ ( 1 + $\frac{d\text{'}}{f_2\mathrm{}}$ ) = $\frac{10}{2.5}$ ( 1 + $\frac{25}{6.25}$ ) = 20

Hence the magnifying power of the microscope is 20

The final image is formed at infinity.

Hence, $v_2$ = $\infty$

Let the object distance for the eyepiece be = $u_2$

According to lens formula, we get

$\frac{1}{v_2}$ - $\frac{1}{u_2}$ = $\frac{1}{f_2}$ or $\frac{1}{u_2}=\frac{1}{v_2}-\frac{1}{f_2}$ or $\frac{1}{u_2}=\frac{1}{\infty }-\frac{1}{6.25}$

$u_2=-6.25cm$

Image distance for the objective lens, $v_1$ = d + $u_2$ = 15 – 6.25 = 8.75 cm

Let the object distance for the eyepiece be = $u_1$

According to lens formula, we get

$\frac{1}{v_1}$ - $\frac{1}{u_1}$ = $\frac{1}{f_1}$ or $\frac{1}{u_1}=\frac{1}{v_1}-\frac{1}{f_1}$ or $\frac{1}{u_1}=\frac{1}{8.75}-\frac{1}{2}$

$u_1=-2.59cm$

Magnitude for the object distance, $\left|u_1\right|$ = 2.59 cm

The magnifying power of a compound microscope is given by the relation,

m = $\frac{v_1}{\left|u_1\right|}\times$ $\frac{d\text{'}}{\left|u_2\right|}$ = $\frac{8.75}{2.59}\times \frac{25}{6.25}$ = 13.51

Hence the magnifying power of the microscope is 13.51.

9.5 Actual depth of the bulb in water, $d_1$ = 80 cm = 0.8 m

Refractive index of water, $\mu$ = 1.33

In the given figure, i = angle of incidence, r = angle of refraction = 90 $°$

The light source, bulb (B) is placed at the bottom of the tank.

Since the bulb is a point source, the emergent light can be considered as a circle of radius, with radius R = $\frac{AC}{2}$ = OA = OB

Using Snell’s law, we can write the refractive index of water as:

$\mu =\frac{\sin ?r}{\sin ?i}$ = $\frac{\sin ?90°}{\sin ?i}$ = 1.33

i = 48.75 $°$

In Δ OBC, $\tan ?i$ = $\frac{OC}{OB}$

$\tan ?48.75°$ = $\frac{R}{80}$

R = 91.23 cm

Hence, the area of the water surface = $\pi R^2$ = 2.61 $\times {10}^4$ ${cm}^2$ = 2.61 $m^2$

9.9 Size of the object, $h_1$ = 3 cm

Object distance, u = - 14 cm

Focal length of the concave lens, f = - 21 cm

Image distance = v

According to lens formula

$\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ - $\frac{1}{-14}$ = $-\frac{1}{21}$

$\frac{1}{v}$ = $-\frac{1}{14}-\frac{1}{21}$ or v = $-\frac{42}{5}$ = - 8.4 cm

Hence, the image is formed on the other side of the lens, 8.4 cm away from the lens. The negative sign shows that the image is erect and virtual.

The magnification of the image is given as:

m = $\frac{Imageheight\left(h_2\right)}{Objectheight\left(h_1\right)}$ = $\frac{v}{u}$

$\frac{h_2}{h_1}$ = $\frac{-8.4}{-14}$ or $h_2$ = 0.6 $\times 3$ = 1.8 cm

If the object is moved further away from the lens, then the virtual image will move towards the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance.

9.7 Refractive index of glass, $\mu$ = 1.55

Focal length required for the double-convex lenses, f = 20 cm

Let the radius of curvature of one face of the lens be = $R_1$ and the other face be = $R_2$

Let the radius of curvature of the double convex lenses be = R

Then $R_1=R$ and $R_2=-R$

The value of R can be calculated as:

$\frac{1}{f}$ = ( $\mu$ - 1) $\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$

$\frac{1}{20}$ = ( $1.55$ - 1) $\left[\frac{1}{R}+\frac{1}{R}\right]$

0.05 = 0.55 $\times \frac{2}{R}$

R = 22 cm

Hence, the radius of curvature for double-convex lens is 22 cm.

9.8 Object distance, u = 12 cm

Focal length of the convex lens, f = 20 cm

Image distance = v

According to lens formula

$\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ - $\frac{1}{12}$ = $\frac{1}{20}$

$\frac{1}{v}$ = $\frac{1}{20}+\frac{1}{12}$ or v = $\frac{60}{8}$ = 7.5 cm

$\mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e},\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{g}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{e}\mathrm{d}\mathrm{}7.5\mathrm{}\mathrm{c}\mathrm{m}\mathrm{}\mathrm{a}\mathrm{w}\mathrm{a}\mathrm{y}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{o}\mathrm{m}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{s},\mathrm{}\mathrm{t}\mathrm{o}\mathrm{w}\mathrm{a}\mathrm{r}\mathrm{d}\mathrm{s}\mathrm{}\mathrm{i}\mathrm{t}\mathrm{s}\mathrm{}\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}.$

Focal length of the concave lens, f = -16 cm

Image distance = v

According to lens formula

$\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ - $\frac{1}{12}$ = $-\frac{1}{16}$

$\frac{1}{v}$ = $-\frac{1}{16}+\frac{1}{12}$ or v = $\frac{48}{1}$ = 48 cm

$\mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e},\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{g}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{e}\mathrm{d}\mathrm{}48\mathrm{}\mathrm{c}\mathrm{m}\mathrm{}\mathrm{a}\mathrm{w}\mathrm{a}\mathrm{y}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{o}\mathrm{m}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{s},\mathrm{}\mathrm{t}\mathrm{o}\mathrm{w}\mathrm{a}\mathrm{r}\mathrm{d}\mathrm{s}\mathrm{}\mathrm{i}\mathrm{t}\mathrm{s}\mathrm{}\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}.$

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.

9.22 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 $\sin ?i_1=\mu \times \sin ?r_1$ = 1.524 $\times \sin ?19°=0.496$

$\angle i_1$ = 29.75 $°$

Hence the angle of incidence is 29.75^$°$

9.25 Area of the virtual image of each square, A = 6.25 ${mm}^2$

Area of each square,  $A_0$ = 1 ${mm}^2$

Hence the linear magnification of the object can be calculated as:

m = $\sqrt{\frac{A}{A_0}}$ = $\sqrt{\frac{6.25}{1}}$ = 2.5, but m = $\frac{v}{u}$ or v = 2.5u

Focal length of the magnifying glass, f = 10 cm. According to lens formula

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

$\frac{1}{10}$ = $\frac{1}{2.5u}$ - $\frac{1}{u}$ or $\frac{1}{10}=-\frac{1.5}{2.5u}$ . Hence u = -6 cm and v = -15 cm

The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e. 25 cm) of a normal eye. Hence, it cannot be seen by the eyes directly.

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.

9.32 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.

9.29 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, $h_2$ = $\theta \times f_o$ = $\frac{1}{30}\times 140$ =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 $\times h_2$ = 6 $\times 4.7=28.2cm$

9.1 Size of the candle, h = 2.5 cm

Let the image size be = h’

Object distance, u = $-$ 27 cm

Radius of curvature of the concave mirror, R = $-$ 36 cm

Focal length of the concave mirror, f = $\frac{R}{2}$ = $-$ 18 cm

Image distance = v

The image distance can be obtained by using mirror formula: $\frac{1}{f}$ = $\frac{1}{u}$ + $\frac{1}{v}$

$\frac{1}{v}$ = $\frac{1}{f}$ $-$ $\frac{1}{u}=\frac{1}{-18}$ $-\frac{1}{-27}$ = $-\frac{1}{54}$

v = −54 cm

Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image.

The magnification of the image is given as:

m = $\frac{h\text{'}}{h}$ = $-\frac{v}{u}$

h’ = $-\frac{v}{u}$ $\times h$ = $-$ $\frac{-54}{-27}$ $\times 2.5$ = - 5 cm

The height of the candle’s image is 5 cm. The negative sign indicates that the image is inverted and virtual. If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.

9.20 Distance between the object and the image ( screen), D = 90 cm

Distance between two locations of the convex lens, d = 20 cm

Let the focal length of the lens be = f

Focal length is related to D and d as:

f = $\frac{D^2-d^2}{4D}$ = $\frac{{90}^2-{20}^2}{4*90}$ = 21.39 cm

Therefore, the focal length of the convex lens is 21.39 cm

9.24 The maximum possible magnification is obtained when the image is formed at the near point (d = 25 cm)

Image distance, v = -d = -25 cm

Focal length, f = 10 cm

Object distance = u

According the lens formula, we have:

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

$\frac{1}{u}=\frac{1}{v}-\frac{1}{f}$ = $\frac{1}{-25}$ - $\frac{1}{10}$ = - $\frac{7}{50}$