Question

In Fraunhofer diffraction pattern, slit width is $0.2 \text{ mm}$ and screen is at $2 \text{ m}$ away from the lens. If the distance between the first minimum on either side of the central maximum is $1 \text{ cm}$ , the wavelength of light used is

• A. $2000 \text{ \AA}$
• B. $4000 \text{ \AA}$
• C. $5000 \text{ \AA}$
• D. $10000 \text{ \AA}$

Hint:

Central maximum width is exactly double the distance from the center to the first minimum.

Answer 1

The Correct Option is C

Step 1: Concept
In Fraunhofer diffraction, the width of the central maximum (distance between first minima) is $W = \frac{2\lambda D}{d}$.

Step 2: Meaning
Here, $W = 1 \text{ cm} = 10^{-2} \text{ m}$, $D = 2 \text{ m}$, and $d = 0.2 \text{ mm} = 2 \times 10^{-4} \text{ m}$.

Step 3: Analysis
$\lambda = \frac{W \cdot d}{2D} = \frac{10^{-2} \times 2 \times 10^{-4}}{2 \times 2} = \frac{2 \times 10^{-6}}{4}$.
$\lambda = 0.5 \times 10^{-6} \text{ m} = 5000 \times 10^{-10} \text{ m}$.

Step 4: Conclusion
The wavelength of light is $5000 \text{ \AA}$. Final Answer: (C)