Question

Wavelength used in single slit diffraction is $628 \text{ nm}$. Width of slit is $0.2 \text{ mm}$. Find the angular width of central maxima in degree:
(1) $0.36^\circ$ (2) $0.38^\circ$

• A. $0.36^\circ$
• B. $0.38^\circ$
• C. $0.45^\circ$
• D. $0.40^\circ$

Answer 1

The Correct Option is A

Step 1: Use the formula for the angular width of central maxima in single slit diffraction.
Angular width $\theta = \frac{2\lambda}{d}$ (in radians), where $\lambda$ is the wavelength and $d$ is the slit width.

Step 2: Identify the given values.
$\lambda = 628 \text{ nm} = 628 \times 10^{-9} \text{ m}$.
$d = 0.2 \text{ mm} = 0.2 \times 10^{-3} \text{ m}$.

Step 3: Calculate the angular width in radians.
$\theta = \frac{2 \times 628 \times 10^{-9}}{0.2 \times 10^{-3}} = \frac{1256 \times 10^{-9}}{2 \times 10^{-4}} = 628 \times 10^{-5} \text{ rad}$.

Step 4: Convert radians to degrees.
To convert radians to degrees, multiply by $\frac{180}{\pi}$.
$\theta^\circ = 628 \times 10^{-5} \times \frac{180}{3.14159} \approx 0.3598^\circ \approx 0.36^\circ$.

Final Answer: $0.36^\circ$. Correct option is (1).