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Physics · Wave Optics

JEE Main 2024 — 30 January, Shift 1 — Question 36

The diffraction pattern of a light of wavelength 400 nm diffracting from a slit of width 0.2 mm is focused on the focal plane of a convex lens of focal length 100 cm . The width of the 1st 1^{\text {st }} secondary maxima will be :

  1. Option A:

    2 mm

    Correct
  2. Option B:

    2 cm

  3. Option C:

    0.02 mm

  4. Option D:

    0.2 mm

Answer: A

Step-by-step solution

Width of 1st 1^{\text {st }} secondary maxima =λa⋅D=\frac{\lambda}{a} \cdot D

Here a=0.2×10−3 ma=0.2 \times 10^{-3} \mathrm{~m} λ=400×10−9 m\lambda=400 \times 10^{-9} \mathrm{~m}

D=100×10−2D=100 \times 10^{-2} Width of 1st 1^{\text {st }} secondary maxima

=400×10−90.2×10−3×100×10−2=\frac{400 \times 10^{-9}}{0.2 \times 10^{-3}} \times 100 \times 10^{-2}

=2 mm=2 \mathrm{~mm}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
30 January, Shift 1
Subject
Physics
Chapter
Wave Optics
Topic
Diffraction of Light Waves
The diffraction pattern of a light of wavelength 400 nm diffracting… | JEE Main 2024 PYQ with Solution · DhiX AI