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
$d(\sin θ - \sin θ_0) = ±λ$
For right of C.M., $\sin θ_1=\sin θ_0 +\frac{λ}{d}$
$=0.5+\frac{0.5}{10}=0.55\,i.e., θ_1 =33.37°$