Question

The condition for obtaining secondary maxima in the diffraction pattern due to a single slit is (symbols have their usual meaning)

• A. $\mathrm{\alpha }\mathrm{sin\theta }=\mathrm{n\lambda }$
• B. $\mathrm{\alpha }\mathrm{sin\theta }=\left(2\mathrm{n}-1\right)\frac{\mathrm{\lambda }}{2}$
• C. $\mathrm{\alpha }\mathrm{sin\theta }=\left(2\mathrm{n}-1\right)\mathrm{\lambda }$
• D. $\mathrm{\alpha }\mathrm{sin\theta }=\frac{\mathrm{n\lambda }}{2}$

Answer 1

The correct option is B

$\mathrm{\alpha }\mathrm{sin\theta }=\left(2\mathrm{n}-1\right)\frac{\mathrm{\lambda }}{2}$

Light:

1. The light travels in a straight line. But, the light can also bend. So, diffraction is a method we can use to bend light.

Diffraction of light:

1. Diffraction refers to the apparent bending of waves around small obstacles and the spreading out of waves past small openings.

The condition for obtaining secondary maxima in the diffraction pattern due to a single slit:

1. To study this bending of light or diffraction, various experiments can be used. Young's single slit experiment is one of them. In this study, light emitted bent as a result of passing through a small split or opening. By measuring the brightness of the surface that the diffracted light illuminates, we can determine the bend's angle.
2. The figure illustrates this as follows:

1. According to above figure, a small slit of width $\mathrm{\alpha }$ is illuminated by the light rays. Light bends and generates a lot of bright and dark patterns on the wall after facing diffraction at the slit's edges.
   The illuminated area of the wall is referred to as the location of the maximum incidences of diffracted light. However, it is believed that the minima of the diffracted light fall in the area where it is dark.
2. The wavelength of the light is related to the distance at which it is incident using a special formula called diffraction. In comparison to the section where the diffracted light falls at its minimum, the maxima portion has a distinct relationship.

So, the secondary maxima in the diffraction pattern due to a single slit is,

$\mathrm{\alpha }\mathrm{sin\theta }=\left(2\mathrm{n}-1\right)\frac{\mathrm{\lambda }}{2}$

Here, $\mathrm{\alpha }$ is the width of the slit, $\mathrm{n}$ is the order of maxima and $\lambda$ is the wavelength of the incident light.

Hence, option $\mathrm{B}$ is the correct option.