Q109 / 150 · MHT CET 2024 (11 May Shift 1) · Physics
In Fraunhofer diffraction pattern, slitwidth is 0.5 mm and screen is at 2 m away from the lens. If wavelength of light used is $5500 Å$, then the distance between the first minimum on either side of the central maximum is $(\theta$ is small and measured in radian)
- A1.1 mm
- B$\quad 2.2 \mathrm{~mm}$
- Correct answerC4.4 mm
- D5.5 mm
Solution
Distance of $1^{\text {st }}$ minima from central maxima
$y_{1 d}=\frac{\lambda D}{a}$
Distance between two minima on either side of the central maxima is $2 \mathrm{y}_{1 \mathrm{~d}}=\frac{2 \lambda \mathrm{D}}{\mathrm{a}}=\frac{2 \times 5500 \times 10^{-10} \times 2}{0.5 \times 10^{-3}}=4.4 \mathrm{~mm}$
Distance between two minima on either side of the central maxima is $2 \mathrm{y}_{1 \mathrm{~d}}=\frac{2 \lambda \mathrm{D}}{\mathrm{a}}=\frac{2 \times 5500 \times 10^{-10} \times 2}{0.5 \times 10^{-3}}=4.4 \mathrm{~mm}$
Hence, the correct answer is option: 'C'.