NVAMediumJEE 2023Interference (Young's Experiment)

JEE Physics 2023 Question with Solution

Two light waves of wavelengths 800nm800 \, \text{nm} and 600nm600 \, \text{nm} are used in Young's double slit experiment to obtain interference fringes on a screen placed 7m7 \, \text{m} away from the plane of slits. If the two slits are separated by 0.35mm0.35 \, \text{mm}, then the shortest distance from the central bright maximum to the point where the bright fringes of the two wavelengths coincide will be ...... mm\text{mm}.

Answer

Correct answer:48

Step-by-step solution

Standard Method

Given: λ1=800nm\lambda_1 = 800 \, \text{nm}, λ2=600nm\lambda_2 = 600 \, \text{nm}, D=7mD = 7 \, \text{m}, and d=0.35mmd = 0.35 \, \text{mm}.

Find: The shortest distance from the central bright maximum where the bright fringes of the two wavelengths coincide.

For Young's double slit experiment, fringe width is

ω=λDd\omega = \frac{\lambda D}{d}

For the first wavelength,

ω1=λ1Dd=800×109×70.35×103=16mm\omega_1 = \frac{\lambda_1 D}{d} = \frac{800 \times 10^{-9} \times 7}{0.35 \times 10^{-3}} = 16 \, \text{mm}

For the second wavelength,

ω2=λ2Dd=600×109×70.35×103=12mm\omega_2 = \frac{\lambda_2 D}{d} = \frac{600 \times 10^{-9} \times 7}{0.35 \times 10^{-3}} = 12 \, \text{mm}

The first common bright fringe occurs at the least common multiple of the two fringe widths:

LCM(16,12)=48mm\text{LCM}(16, 12) = 48 \, \text{mm}

Therefore, the shortest distance is 4848.

Common mistakes

  • Using the difference or sum of the two fringe widths instead of their least common multiple is incorrect, because coincidence requires both bright fringe positions to match exactly. Use the first common multiple of 16mm16 \, \text{mm} and 12mm12 \, \text{mm}.

  • Forgetting unit conversion for slit separation is incorrect, because 0.35mm0.35 \, \text{mm} must be written as 0.35×103m0.35 \times 10^{-3} \, \text{m} before substitution. Mixed units give a wrong fringe width.

  • Taking wavelength in nanometres directly in the formula is incorrect, because 800nm800 \, \text{nm} and 600nm600 \, \text{nm} must be converted to metres first. Always keep λ\lambda, DD, and dd in consistent SI units while calculating ω\omega.

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