Question:
In a Young's double slit experiment, light of 500 nm is used to produce an interference pattern. When the distance between the slits is 0.05 mm, the angular width (in degree) of the fringes formed on the distance screen is close to
In a Young's double slit experiment, light of 500 nm is used to produce an interference pattern. When the distance between the slits is 0.05 mm, the angular width (in degree) of the fringes formed on the distance screen is close to
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Angular fringe width is \(\theta = \lambda / d\) in radians; convert to degrees.
Updated On: Nov 5, 2026
- \(0.17^\circ\)
- \(1.7^\circ\)
- \(0.57^\circ\)
- \(0.07^\circ\)
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The Correct Option is C
Solution and Explanation
Fringe width \(\beta = \frac{\lambda D}{d}\), so the angular width \(\theta = \frac{\beta}{D} = \frac{\lambda}{d}\).
\(\theta = \frac{500 \times 10^{-9}}{0.05 \times 10^{-3}} = 10^{-2}\) rad.
In degrees: \(10^{-2} \times \frac{180}{\pi} = 0.573^\circ \approx 0.57^\circ\).
Answer: (3) \(0.57^\circ\)
\(\theta = \frac{500 \times 10^{-9}}{0.05 \times 10^{-3}} = 10^{-2}\) rad.
In degrees: \(10^{-2} \times \frac{180}{\pi} = 0.573^\circ \approx 0.57^\circ\).
Answer: (3) \(0.57^\circ\)
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