Q97 / 100 · BITSAT 2024 (Memory Based Paper 2) · Physics

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:

  • Correct answer
    A
    $60^{\circ}$
  • B
    $45^{\circ}$
  • C
    $15^{\circ}$
  • D
    $30^{\circ}$

Solution

Condition for minima is $\mathrm{d} \sin \theta=\mathrm{n} \lambda$ For Ist minimad $\sin \theta=\lambda$ $\sin \theta=\frac{\lambda}{a}=\frac{2}{4}=\frac{1}{2} \quad \therefore \theta=30^{\circ}$ Angular spread $=2 \theta=60^{\circ}$

Hence, the correct answer is option: 'A'.

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