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NEET PHYSICSWave OpticsEasy

Question

In Young's double-slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:

A

half

B

four times

C

one-fourth

D

double

Step-by-Step Solution

In Young's double-slit experiment, the fringe width (β\beta) is given by the formula: β=λDd\beta = \frac{\lambda D}{d} where: λ\lambda = wavelength of the light used DD = distance of the screen from the coherent sources dd = separation between the coherent sources

According to the given conditions, the new separation is d′=d2d' = \frac{d}{2} and the new distance of the screen is D′=2DD' = 2D. The new fringe width β′\beta' will be: β′=λD′d′=λ(2D)d2=4(λDd)=4β\beta' = \frac{\lambda D'}{d'} = \frac{\lambda (2D)}{\frac{d}{2}} = 4 \left(\frac{\lambda D}{d}\right) = 4\beta Therefore, the fringe width becomes four times its original value.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Wave Optics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWave Opticsyoungsdoubleslitexperimentseparationbetween

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