Question

Solve this:

​Q45. Light of wavelength 6000 A falls on a single slit of width 0.1 mm. The second minimum will be formed for the angle of diffraction of

(a) 0.08 radian (b) 0.06 radian

(c) 0.12 radian (d) 0.012 radian

Answer 1

Dear Student,


$$\begin{gathered} a\sin \theta =n\lambda \\ forsmallangle\sin \theta \approx \theta \\ \theta =\frac{2*6000*{10}^{-10}}{{10}^{-4}}=0.012radian \\ optiond \\ Regards \end{gathered}$$