Question

The diffraction pattern of a single slit is shown in figure. The point at which the path difference of the extreme rays is two wavelengths is

• A. Point 1
• B. Point 2
• C. Point 4
• D. Point 5

Answer 1

Answer: (D)

Diffraction of light at a single slit, the pattern obtained consists of a central bright band having alternate dark and weak bands of decreasing intensity on both sides
For nth minima (secondary),
$\sin {\theta }_n=\frac{n\lambda }{a}$
For $nth$ maxima (secondary),
$\sin {\theta }_n=\left(2n+1\right)\frac{\lambda }{2a}$
Thus, point 5 corresponds to path difference of extreme rays equal to $2\lambda$.