(i) Consider a parallel beam of light, incident on a reflecting plane surface such as a plane mirror XY as shown in Figure.

(ii) The incident wavefront is AB and the reflected wavefront is A'B' in the same medium. These wavefronts are perpendicular to the incident rays L, M and reflected rays L', M' respectively.
(iii) By the time point A of the incident wavefront touches the reflecting surface, point B is yet to travel a distance BB' to touch the reflecting surface a B'.
(iv) When point B falls on the reflecting surface at H', point A would have reached A.
(v) This is applicable to all the points on the wavefront. Thus, the reflected wavefront A'B' emanates as a plane wavefront. The two normals Nand N' are considered at the points where the rays Land Mfallon the reflecting surface.
(vi) As reflection happens in the same medium, the speed of light is the same before and after the reflection.
(vii) Hence, the time is taken for the ray to travel from B to B' is the same as the time taken for the ray to travel from A to A'.
(viii) Thus, the distance BB' is equal to the distance AA'; (A~A' = BB').
(a) The incident rays, the reflected rays, and the normal are in the same plane.
(b) Angle of incidence,\(\angle i=\angle NAL={ 90 }^{ o }-\angle NAB=\angle BAB'\) 
Angle of reflection,
∠r= ∠N' B' M' = 900 - ∠N' B' A' = A' B' A' 
(ix) For the two right-angle triangles, ΔABB' and ΔB' A' A', the right angles, ∠B and ∠A' are equal, (∠B and∠A = 900); the two sides, ∠A' and ∠B' are equal, (AA'= BB'); the side AB' is common.
(x) 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.

Reflection does not change wavelength.
Frequency remains unchanged both in reflection and refraction
ஃ λ =  5000 Å. Again \(v=\cfrac { c }{ \lambda } =\cfrac { 3\times { 10 }^{ 8 } }{ 5000\times { 10 }^{ -10 } } Hz\) 
= 6 x 1014 Hz
Again i = rand i + r = 90°.
ஃ i = 45°.

Let polaroids P₁ and P₃ be in a crossed positions.
Let the polaroid P₂ make an angle e with the pass axis of polaroid Pr
Let I₁ be the intensity of polarised light emerging our of P₁ Then intensity of light after passing through P₂ will I₂ = I₁cos²θ
Since P₃ and P, are in crossed position, therefore, the angle made by P₂ with P₃ is \(\left( \cfrac { \pi }{ 2 } -\theta \right) \) 
∴ The intensity of light coming out of P₃ is 
\({ I }_{ 3 }={ I }_{ 2 }{ cos }^{ 2 }\left( \cfrac { \pi }{ 2 } -\theta \right) \) 
or \({ I }_{ 3 }={ I }_{ 2 }{ cos }^{ 2 }{ sin }^{ 2 }\theta =\theta ={ I }_{ 1 }\left( \cfrac { 1 }{ 2 } sin20 \right) ^{ 2 }\) 
If I₀ is the intensity of the unpolarised light falling on P_I' then \({ I }_{ 1 }=\cfrac { { I }_{ o } }{ 2 } \) 
\(\therefore { I }_{ 3 }=\cfrac { { I }_{ o } }{ 2 } \left( \cfrac { 1 }{ 2 } sin20 \right) ^{ 2 }\) 
(i) Minimum outcoming intensity is zero.
(ii) Maximum outcoming intensity is received
when \(\theta =\cfrac { \pi }{ 4 } \) 
\(\therefore \left( { I }_{ 3 } \right) _{ max }=\cfrac { { I }_{ o } }{ 2 } \left( \cfrac { 1 }{ 2 } \right) ^{ 2 }=\cfrac { { I }_{ o } }{ 8 } \)

Given data:
In first case, f_e.= 0.03 m
f₀= 0.04 m , L = 0.2 m
\(m=\cfrac { L }{ { f }_{ o } } \left( 1+\cfrac { 0.25 }{ { f }_{ o } } \right) \) 
For m to be the same in both cases,
= \(\cfrac { L }{ 0.03 } \left( 1+\cfrac { 0.25 }{ 0.04 } \right) \) 
= \(\cfrac { 0.2 }{ 0.04 } \left( 1+\cfrac { 0.25 }{ 0.04 } \right) \) 
= \(\cfrac { L }{ 0.03 } \times \cfrac { 29 }{ 4 } =\cfrac { 0.2 }{ 0.04 } \times \cfrac { 28 }{ 3 } \) 
\(L=\cfrac { 5.6 }{ 29 } =0.193m\)

(a) \({ f }_{ o }=\cfrac { 100 }{ 25 } cm=4cm\) 
\({ f }_{ e }=\cfrac { 100 }{ 20 } cm=5cm\) 
Since f₀ therefore the optical instrument is a compound microscope.
(b) \(m=\cfrac { { { v }_{ o } } }{ { u }_{ o } } .\cfrac { D }{ { f }_{ e } } \)