1. sin⁻¹(1/4) 2. sin⁻¹(2/3) 3. sin⁻¹(1/2) 4. sin⁻¹(3/4)

Answer: 4. sin⁻¹(3/4) Key idea: single-slit minima satisfy asinθ=nλ ; secondary maxima are approximately at asinθ=(n+1/2)λ . First minimum at 30^° : asin30^°=λ⇒ a=2λ . First secondary maximum: asinθ=3/2λ . sinθ=3λ/2×2λ=3/4 , so θ=sin⁻¹3/4 .

Topic: Diffraction due to a single slit· Optics

In a diffraction pattern due to a single slit of width 'a', the first minimum is observed at an angle 30^° when light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of :

1. \(\sin^{-1}\left(\frac{1}{4}\right)\)

2. \(\sin^{-1}\left(\frac{2}{3}\right)\)

3. \(\sin^{-1}\left(\frac{1}{2}\right)\)

4. \(\sin^{-1}\left(\frac{3}{4}\right)\)

Doubtly Asked 30 Oct 2026 NEET UG 2016 Phase 1 Practice test neet Save WhatsApp Report

1 answer

Best answer

Answer: 4. \(\sin^{-1}\left(\frac{3}{4}\right)\)

Key idea: single-slit minima satisfy \(a\sin\theta=n\lambda\); secondary maxima are approximately at \(a\sin\theta=(n+\tfrac12)\lambda\).

First minimum at \(30^\circ\): \(a\sin30^\circ=\lambda\Rightarrow a=2\lambda\).
First secondary maximum: \(a\sin\theta=\tfrac32\lambda\).
\(\sin\theta=\dfrac{3\lambda}{2\times2\lambda}=\dfrac34\), so \(\theta=\sin^{-1}\tfrac34\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

Was this answer helpful?

In a diffraction pattern due to a single slit of width 'a', the first minimum is observed at an angle 30^° when light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of :

1 answer

Your answer