In the Young's double slit experiment, the distance between the slits varies in time as
d(t) = d₀ + a₀ sinωt; where d₀, ω and a₀ are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :

Solution

Physics sir · 3 steps

Fringe width goes as 1/d, so the widest fringes come at the smallest gap and the narrowest at the largest.

1. Fringe width. In YDSE the fringe width is β = (λ D)/d. With λ and D fixed, β is largest when d is smallest and smallest when d is largest.
2. Range of d. Since sin ω t runs from -1 to +1, the slit gap runs from dₘᵢₙ = d₀ - a₀ to dₘₐₓ = d₀ + a₀. So βₘₐₓ = (λ D)/(d₀ - a₀) and βₘᵢₙ = (λ D)/(d₀ + a₀).

   βₘₐₓ - βₘᵢₙ = λ D(1/(d₀ - a₀) - 1/(d₀ + a₀)) = (2λ D a₀)/(d₀² - a₀²)

3. So. The difference between the largest and smallest fringe widths is (2λ D a₀)/(d₀² - a₀²), which is option (b).

The idea to keep. When a quantity varies inversely, its extremes are swapped: the smallest d gives the largest β. Take the exact difference of the two extremes. Do not linearise.

Answer from the NTA final answer key. Solution by PrepWiser.