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Wave Optics question

2025 · 22 Jan · Shift 1 · Q68
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Wave Optics question

2025 · 22 Jan · Shift 1 · Q68

JEE MainPhysicsWave OpticsMCQ+4 / −1
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion-(A) : If Young's double slit experiment is performed in an optically denser medium than air, then the consecutive fringes come closer. Reason-(R) : The speed of light reduces in an optically denser medium than air while its frequency does not change. In the light of the above statements, choose the most appropriate answer from the options given below :
  1. A
    (A) is false but (R) is true
  2. B
    Both (A) and (R) are true but (R) is not the correct explanation of (A)
  3. C
    Both (A) and (R) are true and (R) is the correct explanation of (A)
  4. D
    (A) is true but (R) is false
View written solutionFree

Correct answer: C

  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 of light in the medium,
    • DDD = distance of screen from slits,
    • ddd = slit separation.
  2. Effect of optically denser medium

    In an optically denser medium of refractive index μ\muμ, v=cμv = \frac{c}{\mu}v=μc​ Since frequency does not change on entering another medium, λ′=vf=c/μf=λμ\lambda' = \frac{v}{f} = \frac{c/\mu}{f} = \frac{\lambda}{\mu}λ′=fv​=fc/μ​=μλ​

    So the wavelength decreases in the denser medium.

  3. Effect on fringe width

    Now, β′=λ′Dd=λDμd=βμ\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu}β′=dλ′D​=μdλD​=μβ​

    Since μ>1\mu > 1μ>1 for an optically denser medium, β′<β\beta' < \betaβ′<β

    Therefore, the consecutive fringes come closer.

    Hence, Assertion (A) is true.

  4. Check the Reason (R)

    The reason states that in an optically denser medium, speed of light reduces while frequency remains unchanged.

    This is true.

  5. Is (R) the correct explanation of (A)?

    Yes. Because reduced speed with unchanged frequency implies reduced wavelength, and fringe width depends directly on wavelength: β∝λ\beta \propto \lambdaβ∝λ Therefore fringes come closer.

  6. Option analysis

    • A: (A) false, (R) true →\rightarrow→ Incorrect
    • B: Both true, but (R) not correct explanation →\rightarrow→ Incorrect
    • C: Both true, and (R) is the correct explanation →\rightarrow→ Correct
    • D: (A) true, (R) false →\rightarrow→ Incorrect

Final Answer: Option C.

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