Let I be the intensity of beam I incident on first glass plate. Each plate reflects 25% of light incident on it and transmits 75%.
Therefore,
I₁ = I and I₂ = 25/100I = I/4
I₃ = 75/100 I = 3/4I
I₄ = 25/100 I₃ = ¼ x ¾ I = 3/16 I
I₅= 7/100 I₄= ¾ x 3/16 I = 9/64 I
Amplitude ratio of beams 2 and 5 is
R = \(\surd \) I2/I5 = \(\surd \)I/4 x 64/91 = 4/3
I_min/ I_max = [r-1/r+1]² = [4/3-1 / 4/3+1]² = 1/49 = 1:49

The fringe width in the two cases will be \(\beta\) = D\(\lambda \)/d
And \(\beta\)' = D' \(\lambda \)/d
\(\beta\) - \(\beta\) '  = (D - D' )\(\lambda \)/d
or wavelength \(\lambda \) = (\(\beta\) -\(\beta\)' )d /(D - D')
But D -D'  = 5 x 10^-2 m
And  \(\beta\) - \(\beta\)'  = 3 x 10^-5 m, d = 10^-3 m
\(\lambda \) = 3 x 10^-5 x 10^-3 / 5 x 10^-2 = 6 x 10^-7 m = 6000A

A = 60⁰ ,  \(\delta \)_m = 30⁰
i = e = ¾ A = 450
as A + \(\delta \) = i + e
60 + \(\delta \) = 45 +45
or \(\delta \) = 30⁰
Refractive index,
\(\mu \) = sin a + \(\delta \)m /2/sin A/2 = sin 60⁰+30⁰/2/sin 60⁰/2
= sin 45⁰/sin30⁰ = 1\(\surd 2\) 1/2 = \(\surd 2\) = 1.414

(i) As the lens forms a real iamge, it must be a convex lens.
(ii) From the graph, when u = 20 cm , we have v = 20 cm.
For the convex lens forming a real iamge, u is negative and v and f are positive.
U = -20 cm v = +20cm
Using this lens formula,
1/f = 1/v – 1/u = 1/20 – 1/-20 = 1/10 or f = + 10 cm

Clearly , the fish can see the outside view of the cone with semi vertical angle
But \(\mu \) = 1.sin i_c
or 1/3 = 1/ sin i_c
or sin i_c = 3/4 = 0.75 
\(\theta\)/2  =i_c = sin^-1 (0.75 ) = 48.6⁰

Light from the stars near the horizon reaches the earth obliquely through the atmosphere. Its path changes due to refraction. Frequent atmospheric disturbances change the path of light and cause twinkling of stars. Light from the stars overhead reaches the earth normally. It does not suffer refraction. There is no change in its path. Hence there is no Twinkling effect

For unpolarised light electric vector associated with light, is oscillating randomly in all directions in a plane perpendicular to the direction of propagation of light.
In linearly polarised light oscillating electric vector gets aligned along one direction perpendicular to the direction of propagation of light.
[Under the influence of the electric field of the incident wave, the electrons of the scattering molecules, accelerated parallel to the double arrows, do not radiate energy towards the observer. Hence, the scattered light gets polarized.]

For total internal reflection
45° < i_c
sin 45 > sin i_c
5 > sin i_c
> sin i_c
\(\mu\)> sin i_c
or 

For convex lens u = - a, v = R + d
\({1\over f}={1\over v}-{1\over u}\)
\(={1\over (R+d)}-{1\over(-a)}={1\over (R+d)}+{1\over a}\)
\(\Rightarrow\ \ R=\left(af\over (a-f)\right)-d\)

Ray diagram of a reflecting type telescope
Advantages :
(i) Reflecting telescopes have high resolving power due to a large aperture of mirrors.
(ii) Due to availability of paraboloidal mirror, the image is free from chromatic and spherical aberration.

Here, intensity of central maxima will be maximum and that of other will gradually decrease, on this basis graph will be drawn
For central bright fringe, \(\theta ={ 0 }^{ 0 }\)
For first dark fringe, \(a\sin { \theta } =\pm \lambda \)
\(\Rightarrow\)  \(\sin { \theta } =\pm \frac { \lambda }{ a } \)
If \(\theta\) is small, then \(\sin { \theta } \approx \theta \)
So,    \(\theta =\pm \frac { \lambda }{ a } \)

So, the half angular width of central maximum is
\(\theta \approx \sin { \theta } =\frac { \lambda }{ a } \)​​​​​

(i) Relation between critical angle and refractive index of a material is  \(\mu =\frac { 1 }{ sini_{ e } } \)
where i_e = critical angle
\(\mu =\) refractive index of denser medium W.r.t. rarer medium.
(ii) Yes, critical angle depends on colour of light is associated with wavelength .Smaller the wavelength, higher the refractive index and lower the critical angle and vice-versa Like \(\lambda _{ red }>\lambda _{ violet }\) hence \(\mu _{ red }>\mu _{ violet }\)

Figure shows the formation of image A' B' of a finite object AB by a convex mirror, virtual, erect and diminished,
Now \(\triangle ABP=\triangle A'BP\)
\(\therefore \frac { A^{ ' } }{ B^{ ' } } =\frac { PB^{ ' } }{ PB } \)
Applying the new cartesian sign convention,
\(A^{ ' }B^{ ' }=h_{ 2 },AB=h_{ 1 },PB^{ ' }=v,PB=-u\)
\(\therefore \frac { h_{ 2 } }{ h_{ 1 } } =\frac { v }{ -u } \)
Linear magnification m= \(\frac { h_{ 2 } }{ h_{ 2 } } =-\frac { v }{ u } \) 

Polarization.
Two Uses:
Polaroids can be used in sunglasses, window panes, photographic cameras, 3D movie cameras

Power of convex lens = 1/0.25 = 4D
Power of concave lens = 1/0.20 = -5D
Power of the combination, P = P₁+P₂ = -1D
Nature : Diverging