If light with wavelength 0.50 mm falls on a slit of width 10 mm and at an angle θ° = 30° to its normal. Then angular position of first minima located on right sides of the central Fraunhoffer's diffraction will be at
Answer & explanation
Correct answer: option 1
For first diffraction minima at angle θ
$d\left(\sin \theta-\sin \theta_0\right)= \pm \lambda$
For right of C.M., $\sin \theta_1=\sin \theta_0+\frac{\lambda}{d}$
$=0.5+\frac{0.5}{10}=0.55$ i.e., $\theta_1=33.37^\circ$