Power of thin lens,
P = + 5D
\(\mu_g=15 \)
Focal length of thin lens, in air,
\(f_{\text {air }}=\frac{1}{P}=\frac{1}{5}\) = 0.2m = 20cm
By using lens Maker formula,
\( \frac{1}{f_{\text {air }}} =\left(\mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \)
\(=(1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \frac{1}{f_{\text {air }}} =\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
when lens is dipped in liquied, then focal length in liquid.
\(f_l=-100 \mathrm{~cm}\)
Using lens Maker's formula,
\(\frac{1}{f_l}=\left(h \mathrm{k}_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
\( =\left(\frac{\mu_g}{\mu_1}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \)
\(\frac{1}{100}=\left(\frac{15-\mu_l}{\mu_l}\right) \cdot \frac{2}{f_{\text {atr }}} \)
\(\Rightarrow \frac{1}{-100}=\frac{15-\mu_i}{\mu_l} \cdot \frac{2}{20} \)
\(-\frac{1}{10}=\frac{15-\mu_l}{\mu_l} \Rightarrow 15-10 \mu_l=\mu_l \)
\( \Rightarrow \quad \mu_t=\frac{15}{9} \equiv \frac{5}{3} \text { [from Eq. (i)] } \)

In refracting telescope, when the image is formed at infinity, we have
Now, the magnifying power of the telescope,
\(M=\frac{-f_o}{f_e}\) and distance between objective and eye-piece
L = f_o + f_e
According to the question, M=19 and L =1 Then,
\(\Rightarrow \quad 19=\frac{-f_o}{f_e} \text { and } 1=f_o+f_e \Rightarrow f_o=-19 f_e\)
Put value of f₀ in other relation, i.e.,
1 = f_o + f_e
\(\Rightarrow\) -19f_e + f_e = 1 \(\Rightarrow\) -18f_e = 1
\(\Rightarrow \quad f_e=\frac{-1}{18}=-0.0555 \mathrm{~m}=-5.55 \mathrm{~cm}\)
and \(f_o=-19\left(\frac{-1}{18}\right)=\frac{19}{18}=1.0555 \mathrm{~m}\)
= 105.55 cm

By Snell's law,


\(\begin{aligned} \frac{\sin i}{\sin r} & =\frac{n}{\mu} \end{aligned}\)
\(\begin{aligned} \frac{\sin 45^{\circ}}{\sin 90^{\circ}} & =\frac{n}{\mu} \end{aligned}\)
n = \(\mu\) sin 45°            [\(\because\) sin 90° = 1]
\(n=\frac{\mu}{\sqrt{2}}\)                 \(\left[\because \sin 45^{\circ}=\frac{1}{\sqrt{2}}\right]\)
Thus, refractive index ofthe transparent medium is \(n=\frac{\mu}{\sqrt{2}}\)

If the angle of incidence is increased gradually, then the angle of deviation first decreases, attains a minimum value (\(\delta_{m}\)) and then again starts increasing.

When angle of deviation is minimum, the prism is said to be placed in the minimum deviation position.There is only one angle of incidence for which the angle of deviation is minimum.
When \(\delta =\delta_{m}\) [prism in minimum deviation position]
e = i and r₂ = r₁           ....(i)
\(\because\) r₁ + r₂ = A
\(\Rightarrow r+r=A \quad or \quad r=\frac{A}{2}\)
Also, we have
A + \(\delta\) = i + e                ....(ii)
Putting \(\delta\) = \(\delta\)_m and e = i in Eq. (ii), we get
A + \(\delta\)_m = i + i
\(\Rightarrow i=\left ( \frac{A+\delta_{m}}{2} \right )\)
From Snell's law,  \(\mu =\frac{sin \quad i}{sin \quad r}\)
\(\therefore \mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}}\)
This relation is called a prism formula.
For thin prisms (i.e. A is very small), the value of \(\delta\)_m is also very small.
So, \(\begin{aligned} \mu & =\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}} \\ \end{aligned}\) \(\approx \frac{A+\delta_m}{\frac{2}{A / 2}}\)
\(\Rightarrow \quad \delta_m=(\mu-1) A\)

To prove this, suppose O is a point object at an actual depth OA below the free surface of water XY in a tank.

A ray of light incident onXY, normally along OA passes straight along OAA'. Another ray oflight from O incident at ∠i on XY, along OB deviates away from normal. It is refracted at ∠r along BC On producing back, BC meets OA at I. Therefore, I is a virtual image of 0, i.e. when seen through water, O appears at I. Therefore, apparent depth = lA, which is less than the real depth OA.
\( \angle A O B=\angle O B N^{\prime}=i \ \text { (alt. angles) } \)
\(\angle A I B=\angle N B C=r \text { Corresponding angles) } \)
In \(\Delta O A B, \sin i \approx \tan i=\frac{A B}{O B} \quad(\angle i \text { is a small angle. })\)
In \(\Delta I A B, \sin r \approx \tan r=\frac{A B}{I B} \quad(\angle r \text { is a small angle. })\)
Using Snell's law,
n₂ sin i = n₁ sin r
∴ n₂₁ = refractive index of denser medium w.r.t. rarer, i.e. air (n₁ = 1)
\( \frac{n_{2}}{n_{1}} =\frac{\sin r}{\sin i}=\frac{\tan r}{\tan i}=\frac{A B}{A I} \times \frac{A O}{A B} \)
\(\therefore \quad n_{1}=1 ; n_{2}=n \)
\(n =\frac{O A}{I A} \)
\(n =\frac{\text { real depth }}{\text { apparent depth }}=\frac{t}{d}\)

Because refractive index for a given pair of media depends on the ratio of wavelengths and velocity of light in two media but not on frequency, So, frequency remains constant during refraction of light.

(i) Wrong, linear magnification of spherical mirror is \(-\frac{v}{u}\) (using sign conventions). 
(ii) Wrong, as the plane mirror can be considered to be the limit of either a concave or convex spherical curved mirror as the radius, therefore. the focal length of plane mirror becomes infinite.
(iii) Right, but for plane mirror using this formula, focal length becomes infinite.

The incident ray PQ passes through the focus, so the reflected ray is parallel to the principal axis. So, the answer is ray 2.

A convex mirror produces a real image of a virtual object. Therefore, if a beam of light from a virtual object converges to a point behind the convex mirror, then its real image will be formed in front of the mirror.

The image of the object in a plane mirror is as far behind the mirror as the object is in front of it. Therefore, the image of the boy comes near the mirror through the distance equal to that moved by the boy towards the plane mirror. Hence, the image of the boy will approach him with double his speed, i.e. with 4 m/s.

The reflected ray turns twice the angle through which mirror is turned, i.e. 30°.

The distance of normal vision is 25 cm. So if a book is at u = -25 cm, its image should be formed at v = -50 cm. Therefore, the desired focal length is given by
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
\(\text { or } \frac{1}{f}=\frac{1}{-50}-\frac{1}{-25}=\frac{1}{50}\)
or f = + 50 cm (convex lens).

Distance between the image (screen) and the object, D = 90 cm
Distance between two locations of the convex lens, d = 20 cm
Focal length of the lens = f
Focal length is related to d and D as:
f = \(\frac { { D }^{ 2 }-{ d }^{ 2 } }{ 4D } \)
= \(\frac { { (90) }^{ 2 }-({ 20) }^{ 2 } }{ 4\times 90 } =\frac { 770 }{ 36 } =21.3\) cm
Therefore, the focal length of the convex lens is 21.39 cm.

The ray diagram for the formation of the image of the phone is shown in Fig. The image of the part which is on the plane perpendicular to principal axis will be on the same plane. It will be of the same size, i.e., B'C = BC. You can yourself realise why the image is distorted.

You may think that the image will now show only half of the object, but taking the laws of reflection to be true for all points of the remaining part of the mirror, the image will be that of the whole object. However, as the area of the reflecting surface has been reduced, the intensity of the image will be low (in this case, half).

The focal length f = -15/2 cm = -7.5 cm
(i) The object distance u = -10 cm. Then Eq gives
\(\frac { 1 }{ v } +\frac { 1 }{ 10 } =\frac { 1 }{ -7.5 } \)
or \(v=\frac { 10\times 7.5 }{ -2.5 } =-30\) cm
The image is 30 cm from the mirror on the same side as the object
Also, magnification m = \(\frac { v }{ u } =-\frac { (-30) }{ (-10) } =-3\)
The image is magnified, real and inverted.
(ii) The object distance u = -5 cm. Then from Eq
\(\frac { 1 }{ v } +\frac { 1 }{ -5 } =\frac { 1 }{ -7.5 } \)
or \(v=\frac { 5\times 7.5 }{ (7.5-5) } =15\) cm
This image is formed at 15 cm behind the mirror. It is a virtual image.
Magnification m = 15 \(-\frac { v }{ u } =-\frac { 15 }{ (-5) } =3\)
The image is magnified, virtual and erect.

Distance between the object and the image, d = 3 m
Maximum focal length of the convex lens = f_max
For real images, the maximum focal length is given as:
f_max = \(\frac{d}{4}\)
= \(\frac{3}{4}\) = 0.75 m
Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.

Actual depth of the pin, d = 15 cm
Apparent dept of the pin = d'
Refractive index of glass, μ = 1.5
Ratio of actual depth to the apparent depth is equal to the refractive index of glass, i.e.
μ = \(\frac { d }{ { d }^{ ' } } \)
∴ d' = \(\frac { d }{ \mu } \)
= \(\frac{15}{1.5}\) = 10 cm
The distance at which the pin appears to be raised = d' - d
= 15 - 10 = 5 cm
For a small angle of incidence, this distance does not depend upon the location of the slab.

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 is given as:
m = \(\frac{f_o}{f_c}\)
= \(\frac{144}{6}\) = 24
The separation between the objective lens and the eyepiece is calculated as:
f_o + f_e
= 144 + 6 = 150 cm
Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150 cm.

Given, focal length of convex lens, f₁ = 30 cm
Focal length of the concave lens, f₂ = -20 cm
Using the formula of combination of lenses,
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{30}-\frac{1}{20}=\frac{2-3}{60}=-\frac{1}{60}\)
\(\Rightarrow\) f = -60 cm
Since, the focal length of combination is negative in nature. so, the combination behaves like a diverging lens, i.e. as a concave lens.

Refractive index of glass, μ
Focal length of the double-convex lens, f = 20 cm
Radius of curvature of one face of the lens = R₁
Radius of curvature of the other face of the lens = R₂
Radius of curvature of the double-convex lens = R The value of R can be calculated as:
∴ R₁ = R and R₂ = -R
The value of R can be calculated as:
\(\frac { 1 }{ f } =(\mu -1)\left[ \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right] \)
\(\frac { 1 }{ 20 } =(1.55)\left[ \frac { 1 }{ R } +\frac { 1 }{ R } \right] \)
\(\frac { 1 }{ 20 } =0.55\times \frac { 2 }{ R } \)
∴ R = 0.55 x 2 x 20 = 22 cm
Hence, the radius of curvature of the double-convex lens is 22 cm.