Question

Explain the reflection of light at a plane surface with the help of a neat ray diagram. 

Answer 1

 

Reflection of light

XY: Plane reflecting surface
AB: Plane wavefront
RB₁: Reflecting wavefront
A₁M, B₁N: Normal to the plane 
∠AA₁M = ∠BB₁N = ∠i = Angle of incidence
∠TA₁M = ∠QB₁N = ∠r = Angle of reflection

Explanation:

i. A plane wavefront AB is advancing obliquely towards the plane reflecting surface XY. AA₁ and BB₁ are incident rays.

ii. When ‘A’ reaches XY at A₁, then the ray at ‘B’ reaches point ‘P’ and it has to cover distance PB₁ to reach the reflecting surface XY.

iii. Let ‘t’ be the time required to cover distance PB₁. During this time interval, secondary wavelets are emitted from A₁ and will spread over a hemisphere of radius A₁R, in the same medium. Distance covered by secondary wavelets to reach from A₁ to R in time t is the same as the distance covered by primary waves to reach from P to B₁. Thus A₁R = PB₁ = ct.

iv. All other rays between AA₁ and BB₁ will reach XY after A₁ and before B₁. Hence, they will also emit secondary wavelets of decreasing radii.

v. The surface touching all such hemispheres is RB₁ which is reflected wavefront, bounded by reflected rays A₁R and B₁Q.   

vi. Draw A₁M ⊥ XY and B₁N ⊥ XY.  
Thus, angle of incidence is ∠AA₁M = ∠BB₁N = i and angle of reflection is ∠MA₁R = ∠NB₁Q = r. 
∴ ∠RA₁B₁ = 90 - r and ∠PB₁A₁ = 90 - i 

vii. In ΔA₁RB₁ and ΔA₁PB₁ 

∠A₁RB₁ ≅ ∠A₁PB₁  

A₁R = PB₁ (Reflected waves travel equal distance in the same medium in equal time).

A₁B₁ = A₁B₁ ….(common side)  

∴ ΔA₁RB₁ ≅ ΔA₁PB₁ 

∴ ∠RA₁B₁ = ∠PB₁A₁

∴ 90 - r = 90 - i

∴ i = r  

viii. Also, from the figure, it is clear that incident ray, reflected ray, and normal lie in the same plane.

ix. Assuming rays AA₁ and BB₁ to be coming from extremities of the object, A₁B₁ is the size of the object. Distance between corresponding reflected rays A₁T and B₁Q will be the same as A₁B₁ as they are corresponding parts of congruent triangles. This implies the size of the object in a reflected image is the same as the actual size of an object.

x. Also, taking A and B to be right and left sides of the object respectively, after reflection right side at A is seen at T and left side at B is seen at Q. This explains lateral inversion.