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Wave Optics question

2022 · 26 Jul · Shift 1 · Q48
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Wave Optics question

2022 · 26 Jul · Shift 1 · Q48

JEE MainPhysicsWave OpticsMCQ+4 / −1
In Young's double slit experiment, the fringe width is 12 mm12 \mathrm{~mm}12 mm. If the entire arrangement is placed in water of refractive index 43\frac{4}{3}34​, then the fringe width becomes (in mm):
  1. A
    16
  2. B
    9
  3. C
    48
  4. D
    12
View written solutionFree

Correct answer: B

  1. Fringe width in Young's double slit experiment

    The fringe width is given by β=λDd\beta = \frac{\lambda D}{d}β=dλD​ where:

    • λ\lambdaλ = wavelength in the medium
    • DDD = distance between slit and screen
    • ddd = slit separation

  2. Effect of putting the setup in water

    In a medium of refractive index μ\muμ, the wavelength becomes λ′=λμ\lambda' = \frac{\lambda}{\mu}λ′=μλ​

    Therefore, the new fringe width is β′=λ′Dd=λDμd=βμ\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu}β′=dλ′D​=μdλD​=μβ​

  3. Substitute the given values

    Given: β=12 mm,μ=43\beta = 12\,\text{mm}, \qquad \mu = \frac{4}{3}β=12mm,μ=34​

    So, β′=124/3=12×34=9 mm\beta' = \frac{12}{4/3} = 12 \times \frac{3}{4} = 9\,\text{mm}β′=4/312​=12×43​=9mm

  4. Match with options

    9 mm9\,\text{mm}9mm corresponds to Option B.

  5. Comparison with stored answer

    Stored correct answer: B

    Derived answer: B

    Hence, they agree.

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