Skip to content
Question 39 of 46

Q.In a single slit diffraction, how does the angular width of the central maximum change when —

(a) slit width is decreased?
(b) distance between the slit and the screen is increased?
(c) light of smaller visible wavelength is used? Justify your answer in each case. [1+1+1] OR
(a) Define resolving power of a telescope. [1]
(b) How does the resolving power of the instrument depend on the diameter of the objective and the wavelength of the light used? [1+1]
West Bengal WbchseWest Bengal HS (WBCHSE) Board 2024Subjective· 3mImportance★★★★★
85% · 39/46 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The angular width of the central diffraction maximum, 2λ/a2\lambda/a, depends only on wavelength and slit width — not on how far away the screen is — so each factor's effect follows directly from this formula.

In single-slit diffraction, the first minima on either side of the central maximum occur at angle θ\theta satisfying asin⁡θ=λa\sin\theta = \lambda, i.e. (for small angles) θ≈λ/a\theta \approx \lambda/a. The central maximum spans from −θ-\theta to +θ+\theta, so its total angular width is

2θ=2λa2\theta = \dfrac{2\lambda}{a}

(a) Slit width aa decreased: Since angular width ∝1/a\propto 1/a, decreasing aa INCREASES the angular width — the central maximum spreads out more (diffraction effects become more pronounced as the slit narrows relative to the wavelength).

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.