Question

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

• A. point $1$
• B. point $2$
• C. point $4$
• D. point $5$

Answer 1

Answer: (D)

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