Question

In a single slit diffraction pattern, (a) the intensity $I$, at a point where the total phase difference between the wavelets from top to bottom of the slit is 66 rad. (b) If this point is $7^0$ away from the central maxima. Find the width of the slit. Given: $\lambda =600nm.$

Answer 1

(a)$I=I_o[\frac{\sin \left(33rad\right)}{33rad}{]}^2=9.2\times {10}^{-4}{I}_o$
(b)$a=\frac{B\lambda }{2\pi \sin \theta }=\frac{\left(66rad\right)600\times {10}^{-9}}{2\pi \sin 7^0}$
$=5.16\times {10}^{-5}m$ or $=0.052mm$ (nearly)