A microwave of wavelength 2.0 cm falls normally on a slit of width...

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:

Options

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

Correct answer

A. $60^{{}^{\circ}}$

Step-by-step 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}}

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