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

The problem involves finding the angular width of the central maximum in a single-slit diffraction pattern. The given parameters are:

• Wavelength of light, \(\lambda = 628 \, \text{nm} = 628 \times 10^{-9} \, \text{m}\)
• Width of the slit, \(a = 0.2 \, \text{mm} = 0.2 \times 10^{-3} \, \text{m}\)

The formula to determine the angular width of the central maximum is:

The angular width of the central maximum is given by:

\(\theta = \dfrac{2\lambda}{a}\)

Substituting the given values:

\(\theta = \dfrac{2 \times 628 \times 10^{-9}}{0.2 \times 10^{-3}}\)

Simplifying the above expression:

\(\theta = \dfrac{1256 \times 10^{-9}}{0.2 \times 10^{-3}} = \dfrac{1256}{0.2} \times 10^{-6}\)

\(\theta = 6280 \times 10^{-6} \, \text{rad}\)

Converting radians to degrees using the relation \(1\, \text{rad} = 57.2958^\circ\):

\(\theta_{\text{degrees}} = 6280 \times 10^{-6} \times 57.2958\)

\(\theta_{\text{degrees}} \approx 0.36^\circ\)

Thus, the angular width of the central maximum is approximately \(0.36^\circ\).

This matches the given option (1) \(0.36^\circ\), which is the correct answer. Other options do not match with our calculation.