Question

In a single slit diffraction (of width $\alpha$) by a monochromatic source of wavelength $\lambda$ the first minimum of the intensity distribution occurs at an angle

• A. $\lambda/a$
• B. $\lambda/2a$
• C. $a/\lambda$
• D. $a/2\lambda$
• E. $\pi/4$

Hint:

First minimum always at \( \lambda/a \).

Answer 1

The Correct Option is A

Concept: Single slit diffraction (first minimum)
For diffraction through a slit of width $a$, the condition for minima is: \[ a \sin\theta = n\lambda \]

Step 1: First minimum
For first minimum: \[ n = 1 \] \[ a \sin\theta = \lambda \]

Step 2: Small angle approximation
For small $\theta$: \[ \sin\theta \approx \theta \]

Step 3: Final expression
\[ \theta = \frac{\lambda}{a} \] Final Answer:
\[ \boxed{\theta = \frac{\lambda}{a}} \] Note:
Smaller slit width $a$ leads to larger diffraction angle.