Question

A light wave of wavelength ‘λ’ is incident on a slit of width ‘d’. The resulting diffraction pattern is observed on a screen at a distance ‘D’.If linear width of the principal maximum is equal to the width of the slit, then the distance D is

• A. \(\frac {2λ^2}{d}\)
• B. \(\frac {d}{λ}\)
• C. \(\frac {d^2}{2λ}\)
• D. \(\frac {2λ}{d}\)

Answer 1

The Correct Option is C

The angular width of the central maximum (θ) can be given by: 
sin(θ) ≈ \(\frac { λ}{d}\)
Using the small-angle approximation, we can further approximate the angular width as: 
θ ≈ \(\frac { λ}{d}\) 
Using trigonometry, we can relate the width of the central maximum on the screen (W) to the angular width (θ) and the distance D: 
W = 2D x tan(θ)
Substituting the approximate value of θ, we have: 
d = 2D x tan\((\frac { λ}{d})\)
Simplifying, we get: 
D = \(\frac { D^2}{2λ}\)
Therefore, the correct option is (C) \(\frac { D^2}{2λ}\), as it represents the distance D.