Question

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?

Short answer

The distance between the center of the diffraction pattern and the second diffraction minimum is 0.527 m.

Step-by-step solution

Identification of given data

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, $D=3.20\text{m}$

Concept/Significance of single slit diffraction

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.

Determination of the distance between the center of the diffraction pattern and the second diffraction minimum

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 $0.527\text{m}$.