A Fraunhofer diffraction pattern due to a single slit of width 0.2 mm is being obtained on a screen placed at a distance of 2 m from the slit. The first minimum lies at 5 mm on either side of the central maximum on the screen. The wavelength of light is:

5000 Å

The correct answer is Option (2) → 5000 Å

Given:

Slit width $a = 0.2 \, mm = 2 \times 10^{-4} \, m$

Screen distance $D = 2 \, m$

First minimum position $y = 5 \, mm = 5 \times 10^{-3} \, m$

Condition for first minimum:

$a \sin \theta = \lambda$

For small angle, $\sin \theta \approx \tan \theta = \frac{y}{D}$

$\lambda = a \cdot \frac{y}{D}$

Substitute values:

$\lambda = (2 \times 10^{-4}) \cdot \frac{5 \times 10^{-3}}{2}$

$\lambda = (2 \times 10^{-4}) \cdot (2.5 \times 10^{-3})$

$\lambda = 5 \times 10^{-7} \, m$

Final Answer: $\lambda = 500 \, nm$