Q130 / 150 · MHT CET 2021 (20 Sep Shift 2) · Physics

In Fraunhofer diffraction pattern, slit width is $0.2 \mathrm{~mm}$ and screen is at $2 \mathrm{~m}$ away from the lens. If wavelength of light used is $5000 Å$ then the distance between the first minimum on either side of the central maximum is $(\theta$ is small and measured in radian)

  • A
    $2 \times 10^{-2} \mathrm{~m}$
  • B
    $10^{-1} \mathrm{~m}$
  • Correct answer
    C
    $10^{-2} \mathrm{~m}$
  • D
    $10^{-3} \mathrm{~m}$

Solution

Distance between the first minima on either side of the central maximum $=\frac{2 \lambda \mathrm{D}}{\mathrm{a}}=\frac{2 \times 5 \times 10^{-7} \times 2}{0.2 \times 10^{-3}}=10^{-2} \mathrm{~m}$

Hence, the correct answer is option: 'C'.

Relevant Questions