In the single slit experiment of diffraction, the aperture of the slit is 2 mm. If a monochromatic light of wavelength 600 nm is incident normally on the slit, the distance between 2nd order minima and the 4th order maxima on the screen placed 2 m away from the slit will be

1.5 mm

The correct answer is Option (4) → 1.5 mm

Given:

Slit width: $a = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m}$

Wavelength: $\lambda = 600 \, \text{nm} = 6 \times 10^{-7} \, \text{m}$

Distance to screen: $L = 2 \, \text{m}$

1. Position of m-th minima in single slit diffraction:

$y_m = \frac{m \lambda L}{a}$

2nd order minima ($m = 2$):

$y_2 = \frac{2 \cdot 6 \times 10^{-7} \cdot 2}{2 \times 10^{-3}} = \frac{2.4 \times 10^{-6}}{2 \times 10^{-3}} = 1.2 \times 10^{-3} \, \text{m} = 1.2 \, \text{mm}$

2. Position of p-th order maxima (approximate, between minima):

For single slit, maxima are approximately at: $y_p \approx \frac{(p + 0.5) \lambda L}{a}$

4th order maxima ($p = 4$):

$y_4 \approx \frac{(4 + 0.5) \cdot 6 \times 10^{-7} \cdot 2}{2 \times 10^{-3}} = \frac{5.4 \times 10^{-6}}{2 \times 10^{-3}} = 2.7 \times 10^{-3} \, \text{m} = 2.7 \, \text{mm}$

Distance between 2nd minima and 4th maxima:

$\Delta y = y_4 - y_2 = 2.7 \, \text{mm} - 1.2 \, \text{mm} = 1.5 \, \text{mm}$

Answer: $1.5 \, \text{mm}$