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

Answer: (C)

$\frac {d^2}{2λ}$

Answer 2

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.