Question

Monochromatic light of wavelength \(4800 \mathop {\rm{A}}\limits^{\rm{o}}\) falls on slit of width ' \(a\) '. For what value of ' \(a\) ', the first maximum falls at \(30^{\circ} .\left(\sin 30^{\circ}=0.5\right)\)

• A. \(14200\overset{\mathrm{o}}{\mathrm{A}}\)
• B. \(14400\overset{\mathrm{o}}{\mathrm{A}}\)
• C. \(14600\overset{\mathrm{o}}{\mathrm{A}}\)
• D. \(15000\overset{\mathrm{o}}{\mathrm{A}}\)

Hint:

For a single slit, the first secondary maximum occurs where the path difference is three half wavelengths.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
In single slit diffraction, secondary maxima occur at \(a\sin\theta = \left(n+\tfrac12\right)\lambda\). The first maximum has \(n = 1\), so \(a\sin\theta = \dfrac{3\lambda}{2}\).

Step 2: Substitute
\(\lambda = 4800\ \text{\AA}\), \(\theta = 30^{\circ}\), \(\sin 30^{\circ} = 0.5\)
\(a = \dfrac{3\lambda}{2\sin\theta} = \dfrac{3\times 4800}{2\times 0.5}\)

Step 3: Calculation
\(a = 3\times 4800 = 14400\ \text{\AA}\)

Step 4: Compare the options
14400 Å is option B.

Final Answer:
The slit width is 14400 angstrom. \[ \boxed{\text{(B) }14400\,\text{\AA}} \]