\(\frac{1}{f}  =(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)  \)
\(=(1.5-1)\left(\frac{1}{\infty}-\frac{1}{(-10)}\right)\)
(Since plano convex lens)
\(=0.5\left[\frac{1}{10}\right]=\frac{1}{20}  \)
\(\mathrm{f}_t  =20 \mathrm{~cm}\)
Formula for silvered lenses
\(\frac{1}{\mathrm{~F}}  =\frac{2}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_m}  \)
\(\frac{1}{\mathrm{~F}} =\frac{2}{20}+\frac{1}{\infty}  \)
\(\therefore \mathrm{F}  =\frac{20}{2}=10 \mathrm{~cm}\)

\(\frac{I}{f}=\left(\frac{\mu_{\mathrm{L}}}{\mu_L}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
When the biconvex lens of glass dipped in liquid, it acts as a plane sheet of glass.
\(\therefore \mathrm{f}=\infty, \frac{1}{\mathrm{f}}=0 \quad \frac{\mu_g}{\mu_{\mathrm{L}}}-1=0 ; \frac{\mu_{\mathrm{s}}}{\mu_{\mathrm{L}}}=1, \mu_{\mathrm{s}}=\mu_{\mathrm{L}}\)

(i) It produces plane polarised light and functions as a polariser.
(ii) It can also be used to analyse the plane polarised light i.e used at an analyser.

The angular separation between the two extreme colours (violet and red) in the spectrum is called the angular dispersion.

⋋₂ = 600 x 10^-9 m, ⇒ d = 1 x 10^-3 m, D =2m
n⋋ = dsinθ
For first dark fringe
n = 1 For minimum
d sinθ = ⋋, here
CO = Nc (approximately)

From Fig, sinθ \(=\frac{x/2}{D} =x/2D\)
\(sin \theta =\frac{ \lambda}{d} \)
\(\lambda/d=x/2D\)
\(\therefore x =\frac{2D\lambda}{d}=\frac{2 \times 2 \times 600\times 10^{-9}}{ 10^{-3}} \)
\(x=2.4 \times10^{-3}m =2.4\mathrm{~mm} \)

Prisms can be designed to reflect light by 90^o or by 180^o by making use of total internal reflection from the Figures (a) and (b). In the first two cases, the critical angle i_c for the material of the prism must be less than 45^o. Prisms are also used to invert images without changing their size as shown in Figure(c).

Mirage:
Mirage is an optical illusion caused by atmospheric conditions especially the appearance of sheet of water in a desert caused by total internal reflection (or) refraction of light from the sky by heated air.
Looming:
Looming is an optical illusion caused by bending of light which appear an object floating high above its actual position specially in polar region.

The full size of the image will be obtained. But the intensity of image will be reduced. This is because the number of rays of light refracted through different parts of the lens will be reduced.
.

From the lens equation
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ -f } =\cfrac { 1 }{ f } \) 
or \(\cfrac { 1 }{ v } =0\)
∴v = ∞
So, the image is formed at infinity.

From figure,
D = u + v
d = r - u
D + d = u + v + v - u
D + d = 2v

\(v=\frac{D+d}{2}\)
D - d = u + v - v + u = 2u
\(v=\frac{D-d}{2}\)
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \)
\(\frac{1}{f} =\frac{1}{\frac{D+d}{2}}-\frac{1}{\frac{D-d}{2}} \)
\(=\frac{2}{D+d}-\frac{2}{D-d} \)
\(=\frac{2[D-d+D+d]}{D^{2}-d^{2}} \)
\(=\frac{2 \times 2 D}{D^{2}-d^{2}} \)
\(\frac{1}{f} =\frac{4 D}{D^{2}-d^{2}} \quad \therefore f=\frac{D^{2}-d^{2}}{4 D} \)

According to law of reflection,
(i) The incident ray, reflected ray and normal to the reflecting surface all are coplanar (ie. lie in the same plane).
(ii) The angle of incidence i is equal to the angle of reflection r.
i = r

(i) If refraction occurs at the first surface
\(\cfrac { { \mu }_{ 1 } }{ { v }_{ 1 } } -\cfrac { 1 }{ u } =\left( \cfrac { { \mu }_{ 1 }-1 }{ R } \right) \) 
If refraction occurs at the second surface, and the image of the first surface acts as an object.
\(\cfrac { { \mu }_{ 2 } }{ v } -\cfrac { { u }_{ 1 } }{ { v }_{ 1 } } =\cfrac { { \mu }_{ 2 }-{ \mu }_{ 1 } }{ -R } \) 
On adding equations (1) and (2), we get
\(\cfrac { { \mu }_{ 2 } }{ v } -\cfrac { 1 }{ u } =\cfrac { 2{ \mu }_{ 1 }-{ \mu }_{ 2 }-1 }{ R } \) 
If rays are coming from infinity, i.e., u = -∞ then v = f
\(\cfrac { { \mu }_{ 2 } }{ f } +\cfrac { 1 }{ \infty } =\cfrac { { 2\mu }_{ 1 }-{ \mu }_{ 2 } }{ R } \) 
\(f=\cfrac { { \mu }_{ 2 }R }{ { 2{ \mu }_{ 1 }-{ \mu }_{ 2 }-1 } } \) 
(ii) If the combination behaves as a diverging system then f <: O. This is possible only when

\(\Rightarrow 2{ \mu }_{ 1 }-{ \mu }_{ 2 }-1<0\) 
\(2{ \mu }_{ 1 }-{ \mu }_{ 2 }+1\)
\(\Rightarrow { \mu }_{ 1 }<\cfrac { \left( { { \mu }_{ 2 }+1 } \right) }{ 2 } \)
(iii) If the combination behaves as a converging lens then> 0. It is possible only when

\(\Rightarrow { 2\mu }_{ 1 }-{ \mu }_{ 2 }-1>0\) 
\(\Rightarrow 2{ \mu }_{ 1 }->{ \mu }_{ 2 }+1\) 
\({ \mu }_{ 1 }>\cfrac { \left( { \mu }_{ 2 }+1 \right) }{ 2 } \)
The nature of the image formed is real.

Formula
We know \({ i }_{ c }=\cfrac { 1 }{ \mu } \)
 \(\mu =\cfrac { 1 }{ { sini }_{ c } } =\cfrac { 1 }{ { sin45 }^{ o } } =\sqrt { 2 } \) 
According to Brewster's law 
\({ i }_{ p }=\mu =\sqrt { 2 } \) 
\(\Rightarrow { i }_{ p }={ tan }^{ -1 }\sqrt { 2 } \) 
= tan^-1(1.414) ≅ 510 40^o
When light is incident at an angle i_p the corresponding angle of refraction 'r' is given by
i_p + r = 90^o
ஃ r = 90^o- (51^o40^') = (38^o 20')

(i) Let us consider a parallel beam of light, incident on a reflecting plane surface such as a plane mirror XY.
(ii) The incident wavefront is AB and the reflected waterfront is A'B'.
(iii) These wavefronts are perpendicular to the incident rays L, M and reflected rays L', M', respectively.

(i) The incident rays, the reflected rays and the normal are in the same plane.
(ii) Angle of incidence, ∠i = ∠NAL = 90°- ∠NAB = ∠BAB'
Angle of reflection ∠r = ∠N'B'M = 90°- ∠N'B'A'= ∠A'B'A
(a) For the two right angle triangles, ∆ABB' and ∆B'A'A, the two right angles, ∠B and ∠A' are equal, (∠B and ∠A' = 90°); the two sides., AA' and BB' are equal, (AA' = BB'); the side AB' is common
(b) Thus the two triangles are congruent. As per the property of congruency, the two angles, ∠BAB' and ∠A'B'A must also be equal.
i = r
Hence, the laws of reflection are proved.

Dispersion: It is splitting of white light into its constituent colours.
(i) Consider a beam of white light passes through a prism; it gets dispersed into its constituent colours as shown in Figure.
(ii) Let \(\delta_{v}, \delta_{R} \) are the angles of deviation for violet and red light. Let n_V and n_R are the refractive indices for the violet and red light respectively
(iii) The refractive index of the material of a prism is given by the equation
\(\mathrm{n}=\frac{\sin \left(\frac{\mathrm{A}+\mathrm{D}}{2}\right)}{\sin (\mathrm{A} / 2)}\)
(iv) Here A is the angle of the prism and D is the angle of minimum deviation. If the angle of prism is small of the order of 10^o, the prism is said to be a small angle prism.
(v) When rays of light pass through such prisms, the angle of deviation also becomes small. If A be the angle of a smitt angle prism and the angle of deviation then the prism formula becomes.

\(n=\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin (A / 2)}\)
For small angles of  \(A\ and \ \delta \)
\(\sin \frac{A+\delta}{A} \approx \frac{A+\delta}{A} \)
\(\sin \frac{A}{2} \approx \frac{A}{2} \)
\(n=\frac{(A+\delta / 2)}{(A / 2)}=\frac{A+\delta}{A}=1+\frac{\delta}{A} \)
Further simplifying, 
\(\frac{\delta}{A} =n-1 \)
\(\delta =(n-1) A \)    .....(1)
(vi) When white light enters the prism, the deviation is different for different colours. Thus, the refractive index is also different for different colours
For Violet colour, \( \delta_{\mathrm{v}}=\left(\mathrm{n}_{\mathrm{v}}-1\right) \mathrm{A} \)   ...(2)
For Red colour, \(\delta_{\mathrm{R}}=\left(\mathrm{n}_{\mathrm{R}}-1\right) \mathrm{A} \)   ....(3)
(vii) As, angle of deviation for violet colour \(\delta_{v}\) is greater the angle of deviation for red colour \(\delta_{\mathrm{R}}\) the refractive index for violet colour n_v is greater than the refractive index for red colour n_R Subtracting \(\delta_{v}\)  from \(\delta_{\mathrm{R}}\) we get
\(\delta_{\mathrm{v}}-\delta_{\mathrm{R}}=\left(\mathrm{n}_{\mathrm{v}}-\mathrm{n}_{\mathrm{R}}\right) \mathrm{A}\) ....(4)
(viii) The term \(\left(\delta_{v}-\delta_{R}\right)\)  is the angular separation between the two extreme colours (violet and red) in the spectrum is called the angular dispersion. If we take \(\delta\) is the angle of deviation for any middly ray (green or yellow) and the corresponding refractive index. Then,
\(\delta=(n-1) A\)    .....(5)
Dispersive power (ω):
It is the ability of the material of the prism to cause dispersion. It is defined as the ratio of the angular dispersion for the extreme colours to the deviation for any mean colour.
Dispersive power
\(\omega=\frac{\text { Angular dispersion }}{\text { Mean deviation }}=\frac{\delta_{v}-\delta_{R}}{\delta}\)  ....(6)
Substituting \(\left(\delta_{\mathrm{v}}-\delta_{\mathrm{R}}\right) \text { and }(\delta) \)
\(\omega=\frac{\mathrm{n}_{\mathrm{v}}-\mathrm{n}_{\mathrm{R}}}{(\mathrm{n}-1)} \)  .......(7)
(ix) Dispersive power is a dimensionless quality It has no unit. Dispersive power is always positive. The dispersive power of a prism depends only on the nature of material of the prism and it is independent of the angle of the prism.

(i) The angle of view for water animals is restricted to twice the critical angle 2i_c. The critical angle for water is 48.6°. Thus the angle of view is 97.2°.
(ii) The radius R of the circular area depends on the depth d from which it is seen and also the refractive indices of the media.
(iii) The radius R of Snell's window can be deduced with the illustration as shown in Figure.

(iv) Light is seen from a point A at a depth 'd'.
(v) From the Snell's law in product form, n₁ sini = n₂ sinr
(vi) The equation for the refraction happening at the point B on the boundary between the two media is,
n₁ sin i_c = n₂ sin90^o    ..(1)
n₁sini_c = n₂ (∵ sin90^o = 1)
\(sin{ i }_{ c }=\cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \)     ...(2)
From the right angle triangle ΔABC,
\({ sini }_{ c }=\cfrac { CB }{ AB } =\cfrac { R }{ \sqrt { { d }^{ 2 }+{ R }^{ 2 } } } \)      ....(3)
Equating the above two equation 
\(\cfrac { R }{ \sqrt { { d }^{ 2 }+{ R }_{ 2 } } } =\cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \) 
Squaring on both sides
\(\cfrac { { R }^{ 2 } }{ { R }^{ 2 }+d^{ 2 } } \left( \cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \right) ^{ 2 }\) 
Taking reciprocal,
\(\cfrac { { R }^{ 2 }+{ d }^{ 2 } }{ { R }^{ 2 } } =\left( \cfrac { { n }_{ 1 } }{ { n }_{ 2 } } \right) ^{ 2 }\) 
On further simplifying
\(1+\cfrac { { d }^{ 2 } }{ { R }^{ 2 } } =\left( \cfrac { { n }_{ 1 } }{ { n }_{ 2 } } \right) ^{ 2 };\cfrac { { d }^{ 2 } }{ { R }^{ 2 } } =\left( \cfrac { { n }_{ 1 } }{ { n }_{ 2 } } \right) ^{ 2 }-1;\) 
\(\cfrac { { d }^{ 2 } }{ { R }^{ 2 } } =\cfrac { { n }_{ 1 }^{ 2 } }{ { n }_{ 1 }^{ 2 } } -1=\cfrac { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } }{ { n }_{ 2 }^{ 2 } } \) 
Again taking reciprocal and rearranging
\(\cfrac { { R }^{ 2 } }{ { d }^{ 2 } } =\cfrac { { { n }_{ 2 }^{ 2 } } }{ { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } { R }^{ 2 }={ d }^{ 2 }\left( \cfrac { { n }_{ 2 }^{ 2 } }{ { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \right) \) 
∴ The radius of illumination is,
\(R=d\sqrt { \cfrac { { n }_{ 2 }^{ 2 } }{ \left( n_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } \right) } } \)   ...(4)
If the rarer medium outside is air, then, n₂ = 1, and we can take n₁ = n
\(R=d\left( \cfrac { 1 }{ \sqrt { { n }^{ 2 }-1 } } \right) \) or \(R=\cfrac { d }{ \sqrt { { n }^{ 2 }-1 } } \)   ....(5)