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JEE Mains · Physics · STD 12 - Wave optics
  • JEE Mains 2024
  • Hard
  • MCQ

A microwave of wavelength \(2.0 \mathrm{~cm}\) falls normally on a slit of w1dth \(4.0 \mathrm{~cm}\). The angular spread of the central maxima of the diffraction pattern obtained on a screen \(1.5 \mathrm{~m}\) away from the slit, will be _______.

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Answer & Solution

Correct Answer

(C) \(60^{\circ}\)

Step-by-step Solution

Detailed explanation

For first minima a \(\sin \theta=\lambda\)

\(\sin \theta=\frac{\lambda}{\mathrm{a}}=\frac{1}{2}\)

\(\theta=30^{\circ}\)

Angular spread \(=60^{\circ}\)

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