Question

In a single slit diffraction experiment, for wavelength '\(λ\)', half angular width of the principal maxima is '\(θ\)'. Also for wavelength of light '\(pλ\)', the half angular width of the principal maxima is '\(qθ\)'. The ratio of the half angular widths of the first secondary maxima in the first case to second case will be

• A. \(p:1\)
• B. \(1:q\)
• C. \(p:q\)
• D. \(q:p\)

Hint:

Half angular width of the principal maximum is \(\theta=\dfrac\lambda a\); the first secondary maximum lies near \(\dfrac{3\lambda}{2a}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
In single slit diffraction, the first minimum is at \(\sin\theta=\dfrac\lambda a\), which gives the half angular width \(\theta\) of the central maximum. The first secondary maximum lies at about \(\dfrac{3\lambda}{2a}\).

Step 2: Key Formula or Approach
So the angle of the first secondary maximum is \(1.5\) times the half angular width of the principal maximum.

Step 3: Detailed Explanation
Case 1: half angular width \(=\theta\), so the first secondary maximum is at \(1.5\theta\).
Case 2: half angular width \(=q\theta\), so the first secondary maximum is at \(1.5\,q\theta\).
\[ \text{Ratio}=\frac{1.5\theta}{1.5\,q\theta}=\frac1q \]

Final Answer:
The ratio is \(1:q\), option (B). \[ \boxed{1:q\ \text{(B)}} \]