Q4E - University Physics with Modern Physics

Question

Light of wavelength 633 nm from a distant source is incident on a slit 0.750 mm wide, and the resulting diffraction pattern is observed on a screen 3.50 m away. What is the distance between the two dark fringes on either side of the central bright fringe?

Step-by-Step Solution

Answer

The distance between two dark fringes one either side of maxima is 5.91 mm

Step 1: Given Data

Diffraction is a process by which a beam of light spreads when passing through a narrow passage or across the edge of an obstacle accompanied by interference between the waveforms.

Step 2: Total number of dark fringes

It is given that,

Wavelength of light source, λ = 633 nm = 633 x 10^-9 m

Width of the slit, a = 0.750 mm = 7.5 x 10^-4 m

Distance between screen and slit, R = 3.5 m

 

Position of dark fringe in one slit experiment is given by:

$y_m=R tan {\theta }_m$

 

For first dark fringe:

$y_1=R tan {\theta }_1$

 

The angle is given by,

 

$$\begin{gathered} \sin {\theta }_1=\frac{m\lambda }{a} \\ \sin {\theta }_1=\frac{\lambda }{a} \\ {\theta }_1={\sin }^{-1}\left(\frac{\lambda }{a}\right) \end{gathered}$$

 

Putting the value of angle in equation of position of first dark fringe:

 

$y_1=R tan \left[{\sin }^{-1}\left(\frac{\lambda }{a}\right)\right]$

 

Distance between first dark fringe on either side of central maxima is given by

 

$$\begin{gathered} L=2y_1 \\ =2*R \tan \left[{\sin }^{-1}\left(\frac{\lambda }{a}\right)\right] \\ =2*3.50*\tan \left[{\sin }^{-1}\left(\frac{633*{10}^{-9}}{0.75*{10}^{-3}}\right)\right] \\ =5.91*{10}^{-3} m \end{gathered}$$

 

Therefore, the distance between two dark fringes one either side of maxima is 5.91 mm.