In the diffraction pattern due to a single slit of width ' a ' with incident light of wavelength ' λ ', at an angle of diffraction ' θ ', the condition for the second maximum is (a) asin θ = 2λ (b) 2λsin θ = 3a (c) 5λsin θ = 3a (d) 2asin θ = 5λ

Answer: (d). Secondary maxima: asin θ = (2n+1)(λ)/(2), n = 1, 2, Second maximum ( n = 2): asin θ = (5λ)/(2), i.e. 2asin θ = 5λ. So the answer is (d).

In the diffraction pattern due to a single slit of width ' a ' with… | MHT CET 2026 (19 May, Shift 1)

Q2MHT CET 2026 · 19 MAY · SHIFT 1Easy

Diffraction at a Single Slit

In the diffraction pattern due to a single slit of width ' $a$ ' with incident light of wavelength ' $\lambda$ ', at an angle of diffraction ' $\theta$ ', the condition for the second maximum is

Show answer and solution

Answer: (d)

Secondary maxima: $a\sin\theta = (2n+1)\dfrac{\lambda}{2}$ , $n = 1, 2, \dots$
Second maximum ( $n = 2$ ): $a\sin\theta = \dfrac{5\lambda}{2}$ , i.e. $2a\sin\theta = 5\lambda$ .
So the answer is (d).

Report a mistake

Diffraction at a Single Slit question from MHT CET 2026 (19 May, Shift 1)

More questions on Diffraction at a Single Slit