Physics Circle home Student Login

Home›Chapters›Wave Optics PYQs›Diffraction at a Single Slit›Question 1

JEE Main 2026 (2 Apr, Shift 1) · Diffraction at a Single Slit

Diffraction at a Single Slit question from JEE Main 2026 (2 Apr, Shift 1)

Q1JEE Main 2026 · 2 APR · SHIFT 1Easy

Diffraction at a Single Slit

In single slit diffraction pattern, the wavelength of light used is 628 nm and slit width is 0.2 mm, the angular width of central maximum is α×10−2\alpha \times 10^{-2} degrees. The value of α\alpha is ________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 36

  1. Angular width of central maximum =2λa= \dfrac{2\lambda}{a}.
  2. 2×628×10−90.2×10−3=6.28×10−3 rad\dfrac{2 \times 628\times10^{-9}}{0.2\times10^{-3}} = 6.28\times10^{-3}\text{ rad}
  3. In degrees: 6.28×10−3×180π=0.36∘=36×10−26.28\times10^{-3} \times \dfrac{180}{\pi} = 0.36^\circ = 36\times10^{-2} degrees.
  4. So α=36\alpha = 36.
Report a mistake

More questions on Diffraction at a Single Slit