In a single slit diffraction experiment, the aperture of the slit is 3 mm. A monochromatic light of wavelength 620 nm is normally incident on the slit. If the distance between the slit and the screen is 1.2 m, then the separation between the first order minima and the third order maxima on one side of the screen is

$6.20 × 10^{-4} m$

The correct answer is Option (3) → $6.20 × 10^{-4} m$

Slit width: $a=3\ \text{mm}=3\times 10^{-3}\ \text{m}$

Wavelength: $\lambda=620\ \text{nm}=620\times 10^{-9}\ \text{m}$

Screen distance: $D=1.2\ \text{m}$

For single slit diffraction:

Position of $m^{th}$ minima: $y_m=\frac{m\lambda D}{a}$

Position of $n^{th}$ secondary maxima: $y'_n=\frac{(2n+1)\lambda D}{2a}$

First minima: $y_1=\frac{1\cdot \lambda D}{a}=\frac{620\times 10^{-9}\cdot 1.2}{3\times 10^{-3}}=2.48\times 10^{-4}\ \text{m}=0.248\ \text{mm}$

Third maxima ($n=3$): $y'_3=\frac{(2\cdot 3+1)\lambda D}{2a}=\frac{7\lambda D}{2a}$

$y'_3=\frac{7\cdot 620\times 10^{-9}\cdot 1.2}{2\cdot 3\times 10^{-3}}=8.68\times 10^{-4}\ \text{m}=0.868\ \text{mm}$

Required separation: $\Delta y=y'_3-y_1=0.868-0.248=0.620\ \text{mm}$