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Physics · Wave Optics

JEE Main 2026 — 21 January, Morning Shift — Question 35

In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness tt and refractive index n(=1.5)\mathrm{n}(=1.5), the central fringe shifts by 0.2 cm . The value of t is ____\_\_\_\_ cm .

  1. Option A:

    8×10−48 \times 10^{-4}

    Correct
  2. Option B:

    6.0×10−36.0 \times 10^{-3}

  3. Option C:

    5.6×10−45.6 \times 10^{-4}

  4. Option D:

    5.0×10−35.0 \times 10^{-3}

Answer: A

Step-by-step solution

dsin⁡θ=(μ−1)t\mathrm{d} \sin \theta=(\mu-1) \mathrm{t} d[xD]=(μ−1)t\mathrm{d}\left[\frac{\mathrm{x}}{\mathrm{D}}\right]=(\mu-1) \mathrm{t}

t=xdD(μ−1)=(0.2)(0.1)50(1.5−1)\begin{aligned} t & =\frac{x d}{D(\mu-1)} & =\frac{(0.2)(0.1)}{50(1.5-1)} \end{aligned}

t=8×10−4 cm\mathrm{t}=8 \times 10^{-4} \mathrm{~cm}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
21 January, Morning Shift
Subject
Physics
Chapter
Wave Optics
Topic
Young's Double Slit Experiment and Its Modifications
In a double slit experiment the distance between the slits is 0.1 cm… | JEE Main 2026 PYQ with Solution · DhiX AI