Physics Circle home Student Login

Home›Chapters›Wave Optics PYQs›Diffraction at a Single Slit›Question 10

MHT CET 2026 (20 Apr, Shift 2) · Diffraction at a Single Slit

Diffraction at a Single Slit question from MHT CET 2026 (20 Apr, Shift 2)

Q10MHT CET 2026 · 20 APR · SHIFT 2Medium

Diffraction at a Single Slit

In the diffraction pattern, the first maximum is at 30∘30^\circ, when a monochromatic light of wavelength λ\lambda is incident on a slit of width aa. For the same wavelength, if the slit width is changed, so that first maximum is at 45∘45^\circ. The slit width is changed by (sin⁡30∘=12,sin⁡45∘=12)\left(\sin 30^\circ = \frac{1}{2}, \sin 45^\circ = \frac{1}{\sqrt{2}}\right)

Show answer and solution

Answer: (a)

  1. First secondary maximum: asin⁡θ=3λ2a\sin\theta = \dfrac{3\lambda}{2}
  2. At 30∘30^\circ: a1=3λ2×12=3λa_1 = \dfrac{3\lambda}{2 \times \frac{1}{2}} = 3\lambda
  3. At 45∘45^\circ: a2=3λ2×12=3λ2a_2 = \dfrac{3\lambda}{2 \times \frac{1}{\sqrt{2}}} = \dfrac{3\lambda}{\sqrt{2}}
  4. Change =3λ−3λ2=3(2−12)λ= 3\lambda - \dfrac{3\lambda}{\sqrt{2}} = 3\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)\lambda
  5. So the answer is (a).
Report a mistake

More questions on Diffraction at a Single Slit