Question

In young’s double-slit experiment, the fringe width observed is β. if the frequency of the source is tripled, then new fringe width will become

Answer 1

Step 1: Given data

Fringe width observed is $\beta$.

The frequency of the source is tripled.

Step 2: Young's double-slit experiment

1. Light waves from a monochromatic light source are irradiated via two slits separated by a distance d in Young's double-slit experiment.
2. At a distance D, a screen is erected parallel to the two slits, such that d A consequent wave of new strength is created when the two waves interfere at a place on the screen.
3. The light's intensity is determined by the path difference between the two waves. As a result, the screen displays a pattern of bright and dark fringes in succession.
4. Bright fringe denotes the point with the most intensity, while dark fringe denotes the point with the lowest intensity.
5. Demonstrating with the help of a diagram

Step 3: Formula for fringe width with the wavelength of the source

$\beta =\frac{\lambda D}{d}[whereD=dis\tan cebetweenthescreenandslits,discons\tan t]$

Step4: Analyzing

$\beta \propto \lambda .......\left(1\right)$

The formula that relates the frequency with the wavelength of the source is given by,

$f=\frac{c}{\lambda }[wheref=frequency,\lambda =wavelengthofthelight,c=speedoflight]$

So, here, the speed of the light in air is constant, so, the frequency becomes inversely proportional to the wavelength of the light. Hence, we get,

$$\begin{gathered} f\propto \frac{1}{\lambda } \\ \lambda \propto \frac{1}{f}....\left(2\right) \end{gathered}$$

Comparing (1) and (2)

$\beta \propto \frac{1}{f}$

Finally, the relation between the fringe width and the frequency of the source is obtained. So, we can say that the fringe width and the frequency of the source are inversely proportional to each other. Now, we need to find the change in the fringe width when the frequency doubles.

$$\begin{gathered} f\text{'}=3f \\ \frac{1}{\beta \text{'}}=3\frac{1}{\beta }\left[f\text{'}=\frac{1}{\beta \text{'}}\right] \\ \beta \text{'}=\frac{\beta }{3} \end{gathered}$$

Hence, the frequency will decrease by a factor of 3 with an increase in frequency by 3 times.