AIIMS 2019 Physics Young's Double Slit Experiment MCQ Question
A light of wavelength 500 nₘ is incident on a Young's double slit. The distance between slits and screen is D = 1.8 m and distance between slits is d = 0.4 mm. If screen moves with a speed 4 m/s, with what speed first maxima will move?
5 mm/s
4 mm/s
3 mm/s
2 mm/s
Correct Answer
Detailed Explanation
The speed of the first maxima on the screen can be determined using the formula for the fringe width (β) in Young's double-slit experiment, given by β = λD/d, where λ is the wavelength (500 nm), D is the distance to the screen (1.8 m), and d is the distance between the slits (0.4 mm). Calculating β gives approximately 2.25 mm, and since the screen moves at 4 m/s, the speed of the first maxima is the same as the speed of the screen, which is 4 mm/s. Other options are incorrect because they do not accurately reflect the relationship between the screen's speed and the position of the maxima, as the maxima will move at the same speed as the screen.
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