We know that sin \(C=\frac{1}{n} \Rightarrow n=\frac{1}{\sin i_{C}}=\frac{1}{\sin 45^{\circ}}=1.414\)
We conclude that for greater value of refractive index, the value of critical angle is lesser, i.e. the total internal reflection will take place. Here n₁ <  n₂, so, the ray will get internally reflected.
For the 1st ray, n₁ = 1.35
\( \frac{\sin e}{\sin 45} =1.45 \)
\(\Rightarrow \quad \sin e =\frac{1.35 \times 1}{\sqrt{2}}=72.6^{\circ} \)

\( \text { Given: } \angle A=60^{\circ}, \angle i=\angle e, \angle i=\frac{3}{4} \angle A=45^{\circ} \text { and } \)
\(\angle r=\frac{\angle A}{2}=30^{\circ} \)
\(\mu=\frac{\sin i}{\sin r}=\frac{2}{\sqrt{2}}=1.414\)

We know that \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
For a convex lens, j > 0 and u < 0
Given that \(0<|u|<f\)
From lens formula,
\(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\left[\frac{1}{|f|}-\frac{1}{|u|}\right]<0\)
\(\Rightarrow v<0, \text { i.e. } v \text { is negative. }\)
This shows that the image is virtual and lies on the same side as that of the object.
As \(\frac{1}{|v|}<\frac{1}{|u|}, \text { we get }|v|>|u| \Rightarrow m=\frac{|v|}{|u|}>1\)
This shows that the image is enlarged.

The refractive index μ of the lens is less than the refractive index of water \(\text { i.e. } \frac{4}{3}>\mu_L>1\)

In compound microscope objective lens has smaller aperture and smaller focal length than the eyepiece, while in telescope, the objective has a larger aperture and larger focal length than the eyepiece.

Given, \(f=\frac{2}{3} R, R_{1}=+R, R_{2}=-R\)
∴ Using lens Maker's formula,
\(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
\( \Rightarrow \frac{3}{2 R}=(\mu-1)\left(\frac{2}{R}\right) \)
\(\Rightarrow \quad \mu-1=\frac{3}{4} \)
\(\Rightarrow \quad \mu=1+\frac{3}{4}=\frac{7}{4} \)

For a plano-convex lens, R₁ = ∞
R₂ = -R, f = 0.3 m = 30 cm
μ = 1.5
Radius of curvature of plano-convex lens, R = ?
Applying lens Maker's formula, \(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
\(\Rightarrow \frac{1}{30}=(\mu-1)\left ( \frac{1}{\infty }-\frac{1}{-R} \right )=\frac{(1.5-1)}{R}\Rightarrow R=15 cm\)

Since, refractive index of each material is different, so the lens will have two different focal lengths, one for each material. Hence, two images will be formed.

When a glass lens (\(\mu\)_g = 1.5) is immersed in water (\(\mu\)_w = 1.33), then focal length of the lens increases but its nature remains unchanged. due to increase in focal length, its power reduces.

Since, prisms can bend the light rays by 90° and 180° by total internal reflection, so they are used in many optical instruments.

The final image, formed by the combination, is coinciding with the object itself. This implies that the rays, from the object, are retracing their path, after refraction from the lens and reflection from the mirror.
The (refracted) rays are, therefore, falling normally on the mirror. It follows that the rays AB, and A'B' when produced, are meeting at the centre of curvature, C of the mirror. Hence, O₂O = 20 cm, i.e. the radius of curvature of the
mirror.
From the figure, we then see that for the convex lens, u = - 25 cm and v = + (30 + 20) cm = + 50 cm. from the focal length of the lens, we have
\( \frac{1}{50}-\frac{1}{(-25)} =\frac{1}{f} \Rightarrow \frac{1}{f}=\frac{1+2}{50} f =\frac{50}{3} \mathrm{~cm}=16.67 \mathrm{~cm} \)

Given: \(f=20 \mathrm{~cm}, \mu=1.5, \mu_{w}=1.33\)
As \( \frac{1}{f^{\prime}}=\left(\frac{1.5-1.33}{1.33}\right)\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right) \)  .........(i)
Also, \(\frac{1}{20}=(1.5-1)\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right) \) .........(ii)
Dividing equation (i) by (ii), we get
\(\frac{20}{f^{\prime}}=\frac{0.17}{1.33 \times 0.5} \Rightarrow f^{\prime}=78.24 \mathrm{~cm}\)

The refractive index of a transparent medium decreases with increase in wavelength of the incident light.
We have \(\mu_{g a}=\frac{v_{a}}{v_{g}}=\frac{3 \times 10^{8}}{2 \times 10^{8}}=\frac{3}{2}=1.5\)
\( \therefore \quad \mu_{g a}=\frac{1}{\sin i_{c}} \)
\(\Rightarrow i_{c}=\sin ^{-1}\left(\frac{1}{\mu_{g a}}\right) \)
\(\Rightarrow \quad i_{c}=\sin ^{-1}\left(\frac{2}{3}\right)=41.8\)

(i) Given, angle of minimum deviation, 8 m = 30°
Angle of prism, A = 60°
By prism formula, reflected index,
\(\mu=\frac{\sin \frac{\delta_{m}+A}{2}}{\sin A / 2}=\frac{\sin \frac{30^{\circ}+60^{\circ}}{2}}{\sin 30^{\circ}}=\frac{\sin 45^{\circ}}{\sin 30^{\circ}}\)
\(=\frac{1}{\sqrt{2}} \times 2=\sqrt{2}\)
Also, \(\mu=\frac{\text { speed of light in vacuum }(c)}{\text { speed of light in prism }(v)}\)
⇒ v = c/μ = (3 x I0⁸) m/s
Hence, speed of light through prism is
(3 x I0⁸ /\(\sqrt {2}\)) m/s
(ii) Critical angle ic is given as
\(\sin i_{c}=\frac{1}{\sqrt{2}} \quad\left[\because \sin i_{c}=\frac{1}{\mu}\right]\)
⇒  i_c = 45°
A = r + i_c = 60°
⇒ r = 60° - 45° = 15
Using Snell's law \(\frac{\sin i}{\sin r}=\sqrt{2}\)
⇒ sin i = \(\sqrt {2}\) sin r = \(\sqrt {2}\) x sin 15°
∴ i = sin -1 (\(\sqrt {2}\) sin 15°)

(i) In refraction, frequency remains same, so
f _{refracted beam} = f _{incident beam}
Also, \(\mu_{21}=\frac{v_{1}}{v_{2}}=\frac{f \lambda_{1}}{f \lambda_{2}}=\frac{\lambda_{1}}{\lambda_{2}}[\because \ v=f \lambda]\)
\( \Rightarrow v_{2}=\frac{v_{1}}{\mu_{21}}=\frac{3 \times 10^{8}}{133}=2.25 \times 10^{8} \mathrm{~ms}^{-1} \)
\(\therefore \ \lambda_{2}=\frac{\lambda_{1}}{\mu_{21}}=\frac{589}{1.33}=442.85 \approx 443 \mathrm{nm} \)
So, wavelength of reflected beam \(\approx\) 443 nm and its speed = 2.25 x 10⁸ ms ^-1
(ii) For a biconvex lens, using lens Maker's formula,
\(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
Here, f = 20 cm, μ = 1.55 ⇒ R₁ = + R and R₂ = - R
We have, \(\frac{1}{f}=(\mu-1) \frac{2}{R}\)
⇒ R = 2 (μ - 1) f = 2 x (1.55 - 1) x 20 = 22 cm
∴ Radius of 22 cm is required.

(a) The ray diagram showing the formation of image is given below:

The object O is placed in the medium having refractive index n1n1 and the ray is incident from that medium to another medium of refractive index n2.
Here, AB is taken as the perpendicular to the principal axis.
Here, all the angles are very very small,
So, \(\tan \angle A O B=\frac{A B}{O B}\)
As the angle is very small, this becomes
\(\Rightarrow \frac{A B}{O B} \cong \angle A O B\)
Now, \(\tan \angle A C B=\frac{A B}{B C}\)
As the angle is very small, this becomes
\(\Rightarrow \frac{A B}{B C} \cong \angle A C B\)
Again, \(\tan \angle A I B=\frac{A B}{B I}\)
As the angle is very small, this becomes
\(\Rightarrow \frac{A B}{B C} \cong \angle A C B\)
Again, \(\tan \angle A I B=\frac{A B}{B I}\)
As the angle is very small, this becomes
\(\Rightarrow \frac{A B}{B I} \cong \angle A I B\)
Now in \(\triangle A O C\)
Here, it is the exterior angle.
Therefore, \(i=\angle A O B+\angle A C B\)
On putting the values from above equations, we get
\(\Rightarrow i=\frac{A B}{O B}+\frac{A B}{B C}\)
Similarly, in \(\triangle A C I\)
\(r=\angle A B C-\angle A I B\)
On putting values from above equations we get,
\(\Rightarrow r=\frac{A B}{B C}+\frac{A B}{B I}\)
Now using snell's law, we have
\(n_1 \sin i=n_2 \sin r\)
For very small angle we have,
\(\Rightarrow n_1 i=n_2 r\)
Now on putting values from equation (i) and (ii), we have
\(\Rightarrow n_1\left(\frac{A B}{O B}+\frac{A B}{B C}\right)=n_2\left(\frac{A B}{B C}+\frac{A B}{B I}\right)\)
On further solving, we get
\(\Rightarrow \frac{n_1}{O B}+\frac{n_2}{B I}=\frac{n_2-n_1}{B C}\)
Now, from the ray diagram, we have
OB = -u
BI = +v
BC = +R
On putting these values on the above equation we get,
\(\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}\)

If the final image is formed at the distance of distinct visioa, the magnifying power of the telescope is given as
\(m=\frac{f_{o}}{-f_{e}}\left(1+\frac{f_{e}}{D}\right)\)
Astronomical telescope in near point adjustment.

For an astronomical telescope, we will select lens of power 10D for eye piece and lens of power 0.5 D for objective because magnifying power of telescope is \(m=\frac{f_{o}}{f_{e}}\)
Resolving power of the telescope:
Given: \( D_{0}=6 \mathrm{~cm}, \lambda =540 \times 10^{-9} \mathrm{~m} \)
\( \text { R.P. } =\frac{D_{o}}{1.22 \lambda}=\frac{6 \times 10^{-2}}{1.22 \times 540 \times 10^{-9}} \)
\(= 0.9 \times 10^{5} \)

A compound microscope consists of two convex lenses. The lens facing the object to be seen is called an objective lens. The lens facing the eye is called an eye lens. Generally, the eye lens is a combination of lenses and is called an eyepiece. The aperture and the focal length of the objective lens are smaller as compared with those of eye lens. The object to be magnified is placed just beyond the focal point of the objective lens which forms its real, magnified and inverted image. This image acts as the object for the eye lens whose position is so adjusted that the final image formed by the eye lens is at the distance of distinct vision as shown in the diagram.
(b) (i) To achieve a large magni fication of small object, the eyepiece and the objective must have short focal lengths.
(ii) If we place our eyes too close to the eyepiece, the area of the pupil of the eye is less than the area of the eye-ring. So, our eyes will not collect much of the light and our field of view will get reduced.

(ii) Given, the emergent ray grazes along the face AC,

e = 90\(\unicode{xb0} \)
μ = \(\sqrt {2}\)
\(\frac{\sin i}{\sin r_{1}}=\mu=\frac{\sin e}{\sin r_{2}}\)
\(\Rightarrow \ \frac{\sin 90^{\circ}}{\sin r_{2}}=\sqrt{2}\)
\(\text { i.e. } \ \sin r_{2}=\frac{1}{\sqrt{2}} \text { or } r_{2}=45^{\circ}\)
⇒ r₁ + r₂ = ∠A = 60\(\unicode{xb0} \)
r₁ = 60 - r₂ = 15\(\unicode{xb0} \)
\(\Rightarrow \ \frac{\sin i}{\sin 15^{\circ}}=\sqrt{2}\)
⇒ i = 21.47\(\unicode{xb0} \)

The magnifying power of a telescope is equal to the ratio of the visual angle subtended at the eye by final image formed at least distance' of distinct vision to the visual angle subtended at naked eye by the object at infinity.
Magnification, m = \(\frac{I}{O}=\frac{v_{0}}{u_{0}}=\frac{f_{0}}{u_{0}}\)
\(\Rightarrow \ \frac{I}{100}=\frac{150 \times 10^{-2}}{3 \times 10^{3}}\)
⇒ I = 5 x 10^-2 m = 5 cm

() According to lens Maker's formula, focal length of a lens is given by
\(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

Since, \(\frac{1}{f}\) = P (Power of lens)
\(\begin{aligned} & \therefore \quad P=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ & \Rightarrow \quad P \propto \mu \end{aligned}\)
Since, refractive index (\(u\)) of the medium decreases with increase of wavelength (λ) of light. Hence, from Eq.(i),
                                         \(P \propto \mu \propto \frac{1}{\lambda}\)
Hence, with increase of wavelength of light, power of lens decreases.
(ii) For convex lens,
       R₁ = R and R₂ = - R
By lens Maker's formula,
\(\begin{aligned} & \frac{1}{f_{\text {air }}}=\left(\mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ & \frac{1}{-20}=(15-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ & \left(\frac{1}{R_1}-\frac{1}{R_2}\right)=-\frac{1}{10} \end{aligned}\)

When the lens is completely immersed in water, then focal length of lens in water is given by
\(\frac{1}{f_w}=\left({ }^\omega \mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
\( \begin{aligned} f_w & =\left(\frac{\mu_g}{\mu_w}-1\right)\left(-\frac{1}{10}\right) \quad \text { ffr } \\ & =\left(\frac{1.5}{4 / 3}-1\right)\left(-\frac{1}{10}\right) \\ & =\left(\frac{4.5-4}{4}\right)\left(-\frac{1}{10}\right)=-\frac{1}{80} \\ \Rightarrow \quad f_w & =-80 \mathrm{~cm} \end{aligned} $$ [from Eq. (i)]\)

(iv) Given, magnification, m =3
since,           \(m=\frac{v}{u}=3 \Rightarrow v=3 u\)
Power, P = 5D
\(f=\frac{1}{5}=0.2=20 \mathrm{~cm}\)
Now, by lens formula,\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
                            \(\begin{gathered} \frac{1}{20}=\frac{1}{3 u}-\frac{1}{4} \\ \Rightarrow \quad u=-\frac{40}{3} \mathrm{~cm}=-1333 \mathrm{~cm} \end{gathered}\)
Distance of the image,
         V = 34 = 3(-133) = - 40 cm





 

(i) (d): Total internal reflection is the basis for following phenomenon:
(a) Sparkling of diamond.
(b) Optical fibre communication.
(c) Instrument used by doctors for endoscopy.
(ii) (d): Total internal reflection (TIR) is the phenomenon that involves the reflection of all the incident light off the boundary. TIR only takes place when both of the following two conditions are met:The light is in the more denser medium and approaching the less denser medium.The angle of incidence is greater than the critical angle.
(iii) (c) : If incidence of angle, i = critical angle e, then angle of refraction, r = \(90^{\circ}\)
(iv) (b): In optical fibres, core is surrounded by cladding, where the refractive index of the material of the core is higher than that of cladding to bound the light rays inside the core.
(v) (b): From Snell's law, \(\sin C={ }_{1} n_{2}=\frac{v_{1}}{v_{2}}\)
where, c = critical angle = 30° and V₁ and V₂ are speed oflight in medium and vacuum, respectively.
We know that, v² = 3 x 108 m s^-l
\(\therefore \quad \sin 30^{\circ}=\frac{v_{1}}{3 \times 10^{8}} \)
\(\Rightarrow \quad v_{1}=3 \times 10^{8} \times \frac{1}{2} \Rightarrow v_{1}=1.5 \times 10^{8} \mathrm{~ms}^{-1} \)

(i) (d): Refractive index ofa medium depends upon nature and temperature of the medium, wavelength of light.
(ii) (a): Here \(v=5 \times 10^{14} \mathrm{~Hz} ; \lambda=450 \times 10^{-9} \mathrm{~m}\)
\(c=3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}\)
Refractive index of the liquid,
\(\mu=\frac{c}{v}=\frac{c}{v \lambda}=\frac{3 \times 10^{8}}{5 \times 10^{14} \times 450 \times 10^{-9}} \)
\(\mu=1.33\)
(iii) (b): Here i = 60° ; \(\mu\)= 1.5
By snell's law, \(\mu=\frac{\sin i}{\sin r}\)
\(\sin r=\frac{\sin i}{\mu}=\frac{\sin 60^{\circ}}{1.5}=\frac{0.866}{1.5} \)
\(\sin r=0.5773 \text { or } r=\sin ^{-1}(0.58)\)
(iv) (c): As object is at the centre of the sphere, the image must be at the centre only.
\(\therefore\) Distance of virtual image from centre of sphere = 6cm.
(v) (c): Speed of light in second medium is different than that in first medium.

(i) (a): \(\sin C=\frac{1}{\mu}=\frac{1}{3 / 2}=\frac{2}{3}=0.6667\)
\(C=\sin ^{-1}(0.6667)=41.8^{\circ}\)
(ii) (c): \(\mu=\frac{1}{\sin C}=\frac{1}{\sin 48.6}=\frac{1}{0.75}=\frac{4}{3}\)
(iii) (c): From \(\mu=\frac{1}{\sin C}, \sin C=\frac{1}{\mu}\)
\((\text { As } \mu_{v}>\mu_{r} \therefore C_{v}\)
The correct alternative may be (c).
(iv) (b): Difference between apparent and real depth of a pond is due to refraction. Other three are due to total internal reflection
(v) (c) :\(\text { As }^{w} \mu_{g}<^{a} \mu_{w}<^{a} \mu_{g} ; \therefore \theta>\theta_{2}>\theta_{1}\)

(i) (b): Each half lens will form an image in the same plane. The optic axes of the lenses are displaced
\(\frac{1}{v}-\frac{1}{(-30)}=\frac{1}{20} ; v=60 \mathrm{~cm}\)
(ii) (a): Here \(f_{1}=20 \mathrm{~cm} ; f_{2}=?\)
F= 80 cm
\(\text { As } \frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{F} \Rightarrow \frac{1}{f_{2}}=\frac{1}{F}-\frac{1}{f_{1}}\)
\(\frac{1}{f_{2}}=\frac{1}{80}-\frac{1}{20}=\frac{-3}{80}\)
\(f_{2}=\frac{-80}{3}=-26.7 \mathrm{~cm}\)
(iii) (b): The bubble behaves libe a diverging lens
(iv) (b): Convex lens is used in magnifying glass.
(v) (d)

(d) Assertion is false but Reason is true.

(d) Assertion is false but Reason is true.
In air or water, a convex lens made of glass behaves as a convergent lens but when it is placed in carbon disulfide, it behaves as a divergent lens. Therefore, when a convergent lens is placed inside a transparent medium having refractive index greater than that of material of lens, it behaves as a divergent lens.