Question

A narrow single slit of width \( d \) is illuminated by white light. If the first minimum for violet light (\( \lambda = 4500 \, {Å} \)) falls at \( \theta = 30^\circ \), the width of the slit \( d \) in microns is (1 micron = \( 10^{-6} \) m):

• A. 0.4
• B. 0.5
• C. 0.3
• D. 0.7
• E. 0.9

Hint:

For a single slit diffraction pattern, the angular position of the first minimum is given by \( d \sin \theta = m \lambda \).

Answer 1

Answer: (E)

Step 1: The condition for the first minimum in the diffraction pattern produced by a single slit is: \[ d \sin \theta = m \lambda \quad {where} \quad m = 1 \, ({first minimum}) \] Step 2: Substitute the given values: 
- \( \lambda = 4500 \, {Å} = 4500 \times 10^{-10} \, {m} \), 
- \( \theta = 30^\circ \). \[ d \sin 30^\circ = 4500 \times 10^{-10} \] 
Step 3: Since \( \sin 30^\circ = 0.5 \), the equation becomes: \[ d \times 0.5 = 4500 \times 10^{-10} \] Step 4: Solving for \( d \): \[ d = \frac{4500 \times 10^{-10}}{0.5} = 9 \times 10^{-6} \, {m} = 0.9 \, \mu {m} \] Thus, the width of the slit is \( 0.9 \, \mu {m} \).