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MCQMediumJEE 2023Interference (Young's Experiment)

JEE Physics 2023 Question with Solution

In a Young’s double slit experiment, two slits are illuminated with light of wavelength 800 nm800 \, \text{nm}. The first minimum is detected at PP. The value of slit separation aa is:

  • A

    0.4 mm0.4 \, \text{mm}

  • B

    0.5 mm0.5 \, \text{mm}

  • C

    0.2 mm0.2 \, \text{mm}

  • D

    0.1 mm0.1 \, \text{mm}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Wavelength is 800 nm800 \, \text{nm}. The first minimum is detected at PP.

Find: The slit separation aa.

For the first minimum in Young’s double slit experiment, the path difference is

Δx=λ2\Delta x = \frac{\lambda}{2}

From geometry,

a=λDΔxa = \frac{\lambda D}{\Delta x}

Substituting the values shown in the solution,

a=800×10−9×5×10−20.5×10−3=0.2 mma = \frac{800 \times 10^{-9} \times 5 \times 10^{-2}}{0.5 \times 10^{-3}} = 0.2 \, \text{mm}

Therefore, the slit separation is 0.2 mm0.2 \, \text{mm}. The correct option is C.

Common mistakes

  • Using the condition for a maximum instead of a minimum is incorrect here. For the first minimum, the path difference is λ/2\lambda/2, not λ\lambda. Always identify whether the fringe is bright or dark before substituting.

  • Mixing units such as nm, cm, mm, and m leads to a wrong value of aa. Convert all quantities to a consistent unit system before calculation, then express the final answer in mm if required.

  • Treating Δx\Delta x as the fringe width instead of the path difference is wrong. In the shown solution, Δx\Delta x is the path difference corresponding to the first minimum. Keep the geometric meaning of each symbol clear.

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