Question

In Fraunhofer diffraction pattern, slit width is 0.3 mm and screen is at 1.5 m away from the lens. If wavelength of light used is \(4500 \mathop {\rm{A}}\limits^{\rm{o}}\), then the distance between the first minimum on either side of the central maximum is [ \(\theta\) is small and measured in radian.]

• A. 1.5 mm
• B. 2.25 mm
• C. 3.25 mm
• D. 4.5 mm

Hint:

The first minimum is at angle lambda over a; double the screen distance for both sides.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
For a single slit the first minimum satisfies \(a\sin\theta=\lambda\), so \(\theta\approx\dfrac{\lambda}{a}\).

Step 2: Position on screen
\(y=D\theta=\dfrac{\lambda D}{a}=\dfrac{4500\times10^{-10}\times1.5}{0.3\times10^{-3}}=2.25\times10^{-3}\ \text{m}\).

Step 3: Both sides
Distance between the first minima on either side \(=2y=4.5\ \text{mm}\).

Step 4: Compare the options
This is option (D).

Final Answer:
The first minima on either side are 4.5 mm apart. \[ \boxed{\text{(D) }4.5\ \text{mm}} \]