Physics - Wave Optics Question with Solution | TestHub

Angle of first order diffraction $\left({\theta }_1\right)=32^{\circ }$. We know that the angle of diffraction for the $n$th order $\left({\theta }_n\right)$ is given by $d\sin {\theta }_n=n\lambda$. For first order diffraction we get $d\sin 32^{\circ }=1\times \lambda$ or, $\lambda =d\sin 32^{\circ }$. Now for second order diffraction, $d\sin {\theta }_2=2\times \lambda$ or, $d\sin {\theta }_2=2\times d\sin 32^{\circ }$ or $\sin {\theta }_2=2\times \sin 32^{\circ }=2\times 0.529=1.06$. Since the sine of any angle cannot be greater than 1 , therefore there is no second order diffraction.