Q125 / 180 · COMEDK 2026 (9 May Shift 1) · Physics
A diffraction pattern due to a single slit of width 0.12mm is obtained with a blue green light of wavelength 500 nm. The angular separation between central maximum and second order secondary maximum of the diffraction pattern is
- A$0.042 \times 10^{-3}\ rad$
- B$1.042 \times 10^{-4}\ rad$
- C$0.042 \times 10^{-2}\ rad$
- Correct answerD$1.042 \times 10^{-2}\ rad$
Solution
Given slit width, $a = 0.12 \text{ mm} = 1.2 \times 10^{-4} \text{ m}$
Wavelength of light, $\lambda = 500 \text{ nm} = 5 \times 10^{-7} \text{ m}$
The angular position of the $n$-th secondary maximum in a single slit diffraction pattern is given by:
$\theta_n = \dfrac{(2n + 1)\lambda}{2a}$
For the second order secondary maximum, $n = 2$:
$\theta_2 = \dfrac{5\lambda}{2a}$
Substituting the given values:
$\theta_2 = \dfrac{5 \times 5 \times 10^{-7}}{2 \times 1.2 \times 10^{-4}}$
$\theta_2 = \dfrac{25 \times 10^{-7}}{2.4 \times 10^{-4}}$
$\theta_2 = 10.416 \times 10^{-3} \text{ rad}$
$\theta_2 = 1.042 \times 10^{-2} \text{ rad}$
Answer: $1.042 \times 10^{-2}\ rad$
Wavelength of light, $\lambda = 500 \text{ nm} = 5 \times 10^{-7} \text{ m}$
The angular position of the $n$-th secondary maximum in a single slit diffraction pattern is given by:
$\theta_n = \dfrac{(2n + 1)\lambda}{2a}$
For the second order secondary maximum, $n = 2$:
$\theta_2 = \dfrac{5\lambda}{2a}$
Substituting the given values:
$\theta_2 = \dfrac{5 \times 5 \times 10^{-7}}{2 \times 1.2 \times 10^{-4}}$
$\theta_2 = \dfrac{25 \times 10^{-7}}{2.4 \times 10^{-4}}$
$\theta_2 = 10.416 \times 10^{-3} \text{ rad}$
$\theta_2 = 1.042 \times 10^{-2} \text{ rad}$
Answer: $1.042 \times 10^{-2}\ rad$
Hence, the correct answer is option: 'D'.