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NEET PHYSICSWave OpticsMedium

Question

In Young's double slit experiment, the slits are 2 mm2\text{ mm} apart and are illuminated by photons of two wavelengths, λ1=12000 A˚\lambda_1 = 12000\text{ \AA} and λ2=10000 A˚\lambda_2 = 10000\text{ \AA}. At what minimum distance from the common central bright fringe on the screen 2 m2\text{ m} from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?

A

8 mm8\text{ mm}

B

6 mm6\text{ mm}

C

4 mm4\text{ mm}

D

3 mm3\text{ mm}

Step-by-Step Solution

Let the nthn^{\text{th}} bright fringe of wavelength λ1\lambda_1 coincide with the mthm^{\text{th}} bright fringe of wavelength λ2\lambda_2. The position of the nthn^{\text{th}} bright fringe in Young's double slit experiment is given by y=nλDdy = \frac{n\lambda D}{d}. Equating the positions for both wavelengths: nλ1Dd=mλ2Dd\frac{n\lambda_1 D}{d} = \frac{m\lambda_2 D}{d}     nλ1=mλ2\implies n\lambda_1 = m\lambda_2     nm=λ2λ1=10000 A˚12000 A˚=56\implies \frac{n}{m} = \frac{\lambda_2}{\lambda_1} = \frac{10000\text{ \AA}}{12000\text{ \AA}} = \frac{5}{6} For the minimum distance from the central maximum, we take the smallest integers satisfying this ratio, which are n=5n = 5 and m=6m = 6. The minimum distance yy is: y=nλ1Dd=5×12000×1010 m×2 m2×103 my = \frac{n\lambda_1 D}{d} = \frac{5 \times 12000 \times 10^{-10}\text{ m} \times 2\text{ m}}{2 \times 10^{-3}\text{ m}} y=5×12000×107 m=60000×107 m=6×103 m=6 mmy = 5 \times 12000 \times 10^{-7}\text{ m} = 60000 \times 10^{-7}\text{ m} = 6 \times 10^{-3}\text{ m} = 6\text{ mm}

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 Opticsyoungsdoubleexperimentilluminatedphotons

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