A single-slit diffraction experiment is set up with light of wavelength 420 nm, incident perpendicularly on a slit of width 5.10 mm. The viewing screen is 3.20 m distant. On the screen, what is the distance between the center of the diffraction pattern and the second diffraction minimum?

The given data can be listed below,

• The wavelength of the light is,$\lambda =420\text{nm}$
• The value of slit width is,$a=5.10\mu m$
• The distance of screen from slits is, $D=3.20\text{m}$

Sending a beam of light, electrons, or other particles through a single slit results in single slit diffraction. The beam diffracts widely in all directions behind a single, extremely thin (equivalent to the beam's wavelength) slit and covers the entire downstream viewing screen.

The diffraction minima of the single slit experiment is satisfied by,

$a\sin \theta =m\lambda$

Here, ais the slit width, is the wavelength of light and m is the order of diffraction.

The angle of diffraction is given by,

$\sin \theta =\frac{y}{D}$

Here, y is the distance between central and minimum diffraction pattern, and D is the distance of screen and slit.

Substitute all the values in the above,the distance between the center of the diffraction pattern and the second diffraction minimum is calculated as,

$$\begin{gathered} \frac{ay}{D}=m\lambda \\ y=\frac{m\lambda D}{a} \\ =\frac{2\times 420\times {10}^{-9}\text{m}\left(3.20\text{m}\right)}{5.10\times {10}^{-6}\text{m}} \\ =0.527\text{m} \end{gathered}$$

Thus, the distance between the center of the diffraction pattern and the second diffraction minimum is $0.527\text{m}$.