Question

A single slit of width a is illuminated by a monochromatic light of wavelength 600 nm. The value of 'a' for which first minimum appears at θ=30° on the screen will be :

• A. 0.6μm
• B. 1.2μm
• C. 1.8μm
• D. 3μm

Answer 1

The Correct Option is B

Understanding the Problem

We are given the condition for the first minimum in single-slit diffraction and need to find the slit width (\(a\)).

Solution

1. Condition for First Minimum:

The condition for the first minimum in single-slit diffraction is:

\( a \sin \theta = \lambda \)

where:

• \( a \) is the slit width
• \( \theta \) is the angle of the minimum
• \( \lambda \) is the wavelength of light

2. Substitute Values:

Given \(\theta = 30^\circ\) and \(\lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m}\), we have:

\( a \sin(30^\circ) = 600 \times 10^{-9} \, \text{m} \)

3. Solve for Slit Width (a):

We know \(\sin(30^\circ) = \frac{1}{2}\).

Substituting:

\( a \left( \frac{1}{2} \right) = 600 \times 10^{-9} \, \text{m} \)

Solving for \(a\):

\( a = 2 \times 600 \times 10^{-9} \, \text{m} = 1200 \times 10^{-9} \, \text{m} \)

4. Convert to Micrometers:

\( a = 1200 \, \text{nm} = 1.2 \, \mu\text{m} \)

Final Answer

The slit width is \( 1.2 \, \mu\text{m} \).