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MCQEasyJEE 2024Interference (Young's Experiment)

JEE Physics 2024 Question with Solution

Monochromatic light of wavelength 500 nm500 \, \text{nm} is used in Young's double-slit experiment. When one slit is covered with a glass plate (refractive index μ=1.5\mu = 1.5), the central maximum shifts by 44 fringes. Find the thickness of the glass plate.

  • A

    2 μm2 \, \mu \text{m}

  • B

    3 μm3 \, \mu \text{m}

  • C

    4 μm4 \, \mu \text{m}

  • D

    5 μm5 \, \mu \text{m}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: wavelength λ=500 nm\lambda = 500 \, \text{nm}, refractive index μ=1.5\mu = 1.5, fringe shift n=4n = 4.

Find: thickness tt of the glass plate.

The optical path difference introduced by the glass plate is

Δx=t(μ−1)\Delta x = t(\mu - 1)

For a shift of 44 bright fringes,

Δx=nλ=4λ\Delta x = n\lambda = 4\lambda

Equating the two expressions,

t(μ−1)=4λt(\mu - 1) = 4\lambda

Substituting μ=1.5\mu = 1.5 and λ=500 nm\lambda = 500 \, \text{nm},

t(1.5−1)=4×500 nmt(1.5 - 1) = 4 \times 500 \, \text{nm}

So,

t(0.5)=2000 nmt(0.5) = 2000 \, \text{nm} t=20000.5=4000 nm=4 μmt = \frac{2000}{0.5} = 4000 \, \text{nm} = 4 \, \mu \text{m}

Therefore, the thickness of the glass plate is 4 μm4 \, \mu \text{m}. The correct option is C.

Using phase shift relation

Given: the central maximum shifts by 44 fringes when a glass plate is introduced in one path.

Find: thickness tt.

The phase difference produced by the glass plate is written as

Δϕ=2πλ t(μ−1)\Delta \phi = \frac{2\pi}{\lambda} \, t(\mu - 1)

A shift of 44 bright fringes corresponds to an optical path difference of

4λ4\lambda

Hence the equivalent condition is

t(μ−1)=4λt(\mu - 1) = 4\lambda

Using μ=1.5\mu = 1.5 and λ=500 nm\lambda = 500 \, \text{nm},

t=4λμ−1=4×5000.5 nmt = \frac{4\lambda}{\mu - 1} = \frac{4 \times 500}{0.5} \, \text{nm}

Therefore,

t=4000 nm=4 μmt = 4000 \, \text{nm} = 4 \, \mu \text{m}

So the correct option is C.

The first approach in the source mixes phase difference and path difference notation, but its final numerical conclusion still gives 4 μm4 \, \mu \text{m}.

Common mistakes

  • Using tμt\mu instead of t(μ−1)t(\mu - 1) for the additional optical path difference. This is wrong because only the extra path relative to air matters. Always use t(μ−1)t(\mu - 1).

  • Taking the shift of 44 fringes as λ/4\lambda/4 or as a phase angle directly. A shift by 44 bright fringes means the optical path difference is 4λ4\lambda, not a fraction of λ\lambda.

  • Forgetting unit conversion at the end. The calculation gives 4000 nm4000 \, \text{nm}, which must be converted to 4 μm4 \, \mu \text{m}.

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