Question

In a single slit diffraction pattern, the distance between the first minimum on the left and the first minimum on the right is $5\mathrm{mm}$. The screen on which the diffraction pattern is displayed is at a distance of $80~\mathrm{cm}$ from the slit. The wavelength is $6000\mathrm{\AA}$. The slit width in (mm) is about

• A. $0.576$
• B. $0.348$
• C. $0.192$
• D. $0.096$

Hint:

Slit width for single-slit diffraction: $w = \frac{D\lambda}{y_1}$, where $y_1$ is the distance from the centre to the first minimum.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The distance between the first minima on either side of the central maximum is $2y_1 = \frac{2D\lambda}{w}$.
Step 2: Detailed Explanation:
Given $2y_1 = 5$ mm, $D = 80$ cm = $0.8$ m, $\lambda = 6000 \times 10^{-10}$ m.
So, $w = \frac{2D\lambda}{2y_1} = \frac{2 \times 0.8 \times 6000 \times 10^{-10}}{5 \times 10^{-3}} = 0.192 \times 10^{-3}$ m = $0.192$ mm.
Step 3: Final Answer:
The slit width is $0.192$ mm.