Question

Show that the angular fringe width of central maxima is twice that of
5
secondary maxima or minima​

Answer 1

Show that the angular fringe width of central maxima is twice that of the first order secondary maxima or minima.

Explanation :

For first maxima

asinθ = λ

Since θ is very small.

Therefore sinθ = θ

so, aθ = λ

tanθ = y / D

θ = y₁ / D

λ = ay₁ / D

y₁ = λD / a

Since, y₁ = y₂

Therefore, y₂ = λD / a

Angular width of central maxima = 2λD / a

Width of secondary maxima = separation between nth and (n+1)th minima

For minima

θₙ = nλ / d

θₙ₊₁ = (n + 1)λ / d

Angular width of secondary maxima = (n + 1)λ / d - nλ / d

                                                            = λ / d

β = Angular width × D

β = λD / d

So, the angular width of the central maxima is twice that of the first order secondary maxima.