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Physics · Wave Optics

JEE Main 2024 — 1 February, Shift 2 — Question 41

A microwave of wavelength 2.0 cm falls normally on a slit of width 4.0 cm . The angular spread of the central maxima of the diffraction pattern obtained on a screen 1.5 m away from the slit, will be:

  1. Option A:

    30∘30^{\circ}

  2. Option B:

    15∘15^{\circ}

  3. Option C:

    60∘60^{\circ}

    Correct
  4. Option D:

    45∘45^{\circ}

Answer: C

Step-by-step solution

For first minima a sin⁡θ=λ\sin \theta=\lambda sin⁡θ=λa=12\sin \theta=\frac{\lambda}{\mathrm{a}}=\frac{1}{2} θ=30∘\theta=30^{\circ} Angular spread =60∘=60^{\circ}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
1 February, Shift 2
Subject
Physics
Chapter
Wave Optics
Topic
Diffraction of Light Waves

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