JEE Mains · Physics · STD 12 - 10. Wave optics
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 _______.
- A \(30^{\circ}\)
- B \(15^{\circ}\)
- C \(60^{\circ}\)
- D \(45^{\circ}\)
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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