The components of electric vector associated with light wave, along the direction of aligned molecules of a polaroid, get absorbed. As a result after passing through it, the components perpendicular to the direction of aligned molecules will be obtained in the form of plane polarised light.
(b) When unpolarised light is incident on the boundary between two transperent media, the reflected light is polarised, with electric vector perpendicular to the plane of incidence when the reflected and refracted light rays make a right angle, as shown in the figure below.
Since, \(\angle CBQ+\angle QBD={ 90 }^{ o }\)
(90 - i_B) + (90 - r) = 90^o
i_B + r = 90
r = 90 - i_B
Using Snell's law,
\(\mu =\frac { sin{ i }_{ B } }{ sinr } \)
\(=\frac { sin{ i }_{ B } }{ sin\left( 90-{ i }_{ B } \right) } \)
\(=\frac { sin{ i }_{ B } }{ { cosi }_{ B } } \)
\(\mu =tan{ i }_{ B }\).

(i) The teacher displays the qualities of deep knowledge of the phenomenon and eagerness to explain it to the child.
(ii) The phenomenon involved in a thin film is interference. Different colours of light interfere at different points in space and hence, child is able to see different colours.

(i) Being a phtsics student, he knows that light rays from inside the glass bends away from the normal and appear to diverge. So, it gives false impression that ther is more amount of liquid in bottle.
(ii) Affection, patience and knowledge about refraction.

For microscope f₀ = 1.25cm, f_e = 5cm
When final image forms at infinity, then mangnification produced by eye lens is given by
\(m=-\cfrac { L }{ f_{ 0 } } .\cfrac { D }{ { f }_{ e } } \)
\(\Rightarrow \)  \(-30=-\cfrac { L }{ 1.25 } \times \cfrac { 25 }{ 5 } \)                                                    
\(L=\cfrac { 30\times 1.25 }{ 5 } \)         
\(\Rightarrow \) \(L=7.50cm\)                                                             
For objective lens,v₀ = L = 7.5 cm, f₀ = 1.25 cm
Applying lens formula,
\(\cfrac { 1 }{ f_{ 0 } } =\cfrac { 1 }{ { v }_{ 0 } } -\cfrac { 1 }{ { u }_{ 0 } } \)
\(\Rightarrow \)  \(\cfrac { 1 }{ 1.25 } =\cfrac { 1 }{ 7.5 } -\cfrac { 1 }{ { u }_{ 0 } } \)                                                               
\(\cfrac { 1 }{ { u }_{ 0 } } =\cfrac { 1 }{ 7.5 } -\cfrac { 1 }{ 1.25 } \)
\(=\cfrac { 1.25-7.5 }{ 7.5\times 1.25 } =-\cfrac { 6.25 }{ 7.5\times 1.25 } \)        
\(\Rightarrow \)  \({ u }_{ 0 }=\cfrac { 7.5\times 1.25 }{ 6.25 } =-1.5cm\)
The object must be at a distance of 1.5 cm from objective lens. 

For convex mirror, f>0, u<0
From mirror formula, \((\frac { 1 }{ v } =\frac { 1 }{ f } -\frac { 1 }{ u } ,so\frac { 1 }{ v } >\frac { 1 }{ f } orv\)
Also \(\frac { 1 }{ v } >\frac { -1 }{ u } or\frac { -v }{ u } <1\quad i.e.\quad m<1\)
Thus, image is always located between pole and focus of the mirror and is always dimished in size.

here, f₀=1 m=100cm, f_e=20cm.
If h₁=size of image formed by objective lens and
h₂=size of image formed by eye piece
\(tan\alpha={h_1\over f_0}={h_1\over 100}\)
\(m_e={h_2\over h_1}=1+{d\over f_e}\)
\({10\over h_1}=1+{24\over14}={44\over20}\ \ h_1={200\over44}={50\over11}cm\)
\(\alpha\simeq tan\ \alpha ={h_1\over100}={50\over1100}={1\over22}rad\)
= 0.0455 rad 

(i) The time elapsed to travel from S to \({ P }_{ 1 }\) 
\(i_{ 1 }=\frac { { SP }_{ 1 } }{ C } =\frac { \sqrt { { u }^{ 2 }+{ b }^{ 2 } } }{ c }\) 
\(or \ \frac { u }{ c } \left( 1+\frac { 1 }{ 2 } \frac { { b }^{ 2 } }{ { u }^{ 2 } } \right) \ assuming \ b<<{ u }_{ o }\)
Time required to travel from \({ p }_{ 1 }\) to  O is 
\(t_{ 2 }=\frac { { p }_{ 1 }o }{ c } =\frac { \sqrt { { v }^{ 2 }+{ b }^{ 2 } } }{ c } =\frac { v }{ c } \left( 1+\frac { 1 }{ 2 } \frac { { b }^{ 2 } }{ { v }^{ 2 } } \right) \) 
Time required to travel through the lens is 
\(t_{ 1 }=\frac { { \left( n-1 \right) \omega (b) } }{ c } \) 
where n is the refractive index
Thus, the total time is
\(t=\frac { { 1 } }{ c } u+v+\frac { 1 }{ 2 } { b }^{ 2 }\left( \frac { 1 }{ u } +\frac { 1 }{ v } \right) +(n-1)w(b)\)
\(put\ \frac { 1 }{ D } =\frac { 1 }{ u } +\frac { 1 }{ v } \)
\( Then, t=\frac { { 1 } }{ c } \left( u+v+\frac { 1 }{ 2 } \frac { { b }^{ 2 } }{ D } +(n-1)\left( { \omega }_{ 0 }+\frac { { b }^{ 2 } }{ \alpha } \right) \right) \)
Fermat's principle gives the time taken should be minimum
For that first derivative should be zero.
\(\frac { dt }{ db } =0=\frac { b }{ CD } -\frac { 2(n-1)b }{ c\alpha }\) 
\( \\ \ \alpha \ =\ 2(n-1)D\)
Thus, a convergent lens is formed if \(\alpha \ =\ 2(n-1)D\) 
This is independent of and hence, all paraxial rays from S will converge at O i.e., for rays and (b<
\(since,\ \frac { 1 }{ D } =\frac { 1 }{ u } +\frac { 1 }{ v } ,\ \) the focal length is D
(ii) In this case, differentiating the expression of time taken t w,r,t,b.
\(t=\frac { 1 }{ c } \left( u+v+\frac { 1 }{ 2 } \frac { { b }^{ 2 } }{ D } +(n-1){ k }_{ 1 }In\left( \frac { { k }_{ 2 } }{ b } \right) \right) \)
\( \frac { dt }{ db } =0=\frac { b }{ D } -(n-1)\frac { { k }_{ 2 } }{ b } \)
\( { b }^{ 2 }=(n-1){ k }_{ 1 }D\)
\(\therefore \quad b=\sqrt { (n-1){ k }_{ 1 }D } \)
Thus, all rays passing at a height b shall contribute to the image. The ray paths make an angle 
\( \beta =\frac { b }{ v } =\sqrt { \frac { (n-1){ k }D }{ { v }^{ 2 } } }\)
\(={ \left[ \frac { (n-1) \ { k }_{ 1 }uv }{ { v }^{ 2 }(u+v) } \right] }^{ 1/2 }\)
\( i.e., \beta ={ \left[ \frac { (n-1)ku }{ (u+v)\theta } \right] }^{ 1/2 }\)

Here ray of light at (x, y) enters at 90°. consider a portion of the ray between x and x+dx inside the liquid. the ray deviates at an angle \(d\theta \) between x  and x+dx. emerging at (x+dx, y+dy) at an angle \(\theta +d\theta \)  while entering at \(\theta \), at height ay and y + dy. from snell's law
\( \mu (y) \ sin \ \theta \ = \ \mu (y+dy) \ sin \ (\theta +d\theta )\)
\( or\ \mu (y) \ sin\ = \ \left( \mu (y)+\frac { d\mu }{ dy } dy \right) (sin \ \theta \ cos \ d\theta +cos \ \theta \ sin \ d\theta )\)

As \(d\theta \) is small, cos \(d\theta \) = 1 and sin \(d\theta \) = \(d\theta \) 
\(\therefore \mu (y) \ sin \ \theta \ = \mu (y) \ sin \ \theta + \ \mu (y) \ cos \ \theta \ d\theta +\frac { d\mu }{ dy } dy \ sin \ \theta \)
( the fourth term is negligibly small hence neglected)
\(or \ \mu (y) \ cos \ \theta d\theta \ =-\frac { d\mu }{ dy } dy \ sin \ \theta \)
\(o d\theta \ =-\frac { 1 }{ \mu } \frac { d\mu }{ dy } dy \ tan \ \theta \)
but from the figure, we find  \(tan \ \theta =\frac { dx }{ dy } \) 
\(\Rightarrow \ dy \ tan \ \theta =dx\)
\(sod\theta =-\frac { 1 }{ \mu } \frac { d\mu }{ dy } dx\)
\( or\theta =-\frac { 1 }{ \mu } \frac { d\mu }{ dy } \int _{ 0 }^{ d }{ dx } =-\frac { 1 }{ \mu } \frac { d\mu }{ dy } (d)\)
This is the deviation in travelling a horizontal distance d.

Differences between interference and diffraction

Interference of Light. The phenomenon of redistribution of energy in a medium due to superimposition of waves from two coherent source of light is called Interference of Light.

Constructive Interference.  At points, where the crest of one wave falls the crest of the other or a through of one falls on the through of the other, the amplitude of the resulting wave becomes maximum. Hence the energy or the intensity of light at such points becomes maximum. This is called Constructive Interference.

Destructive Interference. At some other points where the through of one falls on the crest of the other or crest of one falls on the through of the other, the amplitude of the resulting waves becomes minimum. Hence the energy or intensity becomes minimum. This is called Destructive Interference.

Law of conservation of energy is obeyed. It should be clearly understood that in interference of light no light energy is destroyed. The loss of energy at the points of destructive interference appears as the increase of energy at the points of constructive interference.