Question

The wavelength of a monochromatic light which is used in single slit diffraction is \(800\,\text{nm}\). The width of the single slit for which the first minimum appears at \(\theta = 45^\circ\) on the screen will be:

• A. \(1.13\,\mu m\)
• B. \(1.23\,\mu m\)
• C. \(2.13\,\mu m\)
• D. \(1.3\,\mu m\)

Hint:

For first minimum in single slit diffraction, always use \(a \sin\theta = \lambda\). No factor of 2 or higher order term is used.

Answer 1

The Correct Option is A

Step 1: Use condition for first minimum in single slit diffraction.
For first minimum:
\[ a \sin\theta = \lambda \]
where \(a\) is slit width.

Step 2: Convert wavelength into SI units.
\[ \lambda = 800\,\text{nm} = 800 \times 10^{-9}\,\text{m} \]

Step 3: Substitute given angle.
\[ \sin 45^\circ = \frac{1}{\sqrt{2}} \]
So,
\[ a \times \frac{1}{\sqrt{2}} = 800 \times 10^{-9} \]

Step 4: Solve for slit width.
\[ a = 800 \times 10^{-9} \times \sqrt{2} \]
\[ a = 800 \times 1.414 \times 10^{-9} \]
\[ a \approx 1131 \times 10^{-9}\,\text{m} \]

Step 5: Convert into micrometers.
\[ a \approx 1.13 \times 10^{-6}\,\text{m} \]
\[ a = 1.13\,\mu m \]

Step 6: Final answer.
\[ \boxed{1.13\,\mu m} \]

Step 7: Concept clarity.
The first minimum in single slit diffraction always satisfies \(a \sin\theta = \lambda\), which directly gives slit width when angle and wavelength are known.