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(C) For the first diffraction minima,the condition is given by $a \sin 	heta = n \lambda$. For the first minimum,$n = 1$.Given: wavelength $\lambda =

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$60$

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(C) For the first diffraction minima,the condition is given by $a \sin 	heta = n \lambda$. For the first minimum,$n = 1$.Given: wavelength $\lambda = 2.0 \ cm$ and slit width $a = 4.0 \ cm$.Substituting the values: $\sin 	heta = \frac{\lambda}{a} = \frac{2.0}{4.0} = 0.5$.Thus,$	heta = \arcsin(0.5) = 30^{\circ}$.The angular spread of the central maxima is the angle between the first minima on either side of the central maximum,which is $2	heta$.Therefore,angular spread $= 2 	imes 30^{\circ} = 60^{\circ}$.

$A$ 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: (in $^{\circ}$)

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