Question

In a Young's double slit interference experiment, the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength \[\lambda.\] Find the distance from the central point where the intensity falls to (a) half the maximum, (b) one-fourth the maximum.

Answer 1

Given:-

Separation between the two slits = d

Wavelength of the light = \[\lambda\]

Distance of the screen = D

(a) When the intensity is half the maximum:-

Let I_max be the maximum intensity and I be the intensity at the required point at a distance y from the central point.

So,

\[I =  a^2  +  a^2  + 2 a^2 \cos\phi\]

Here, \[\phi\] is the phase difference in the waves coming from the two slits.

So, \[I = 4 a^2  \cos^2 \left( \frac{\phi}{2} \right)\]

\[\Rightarrow \frac{I}{I_\max} = \frac{1}{2}\]

\[ \Rightarrow \frac{4 a^2 \cos^2 \left( \frac{\phi}{2} \right)}{4 a^2} = \frac{1}{2}\]

\[ \Rightarrow \cos^2 \left( \frac{\phi}{2} \right) = \frac{1}{2}\]

\[ \Rightarrow \cos\left( \frac{\phi}{2} \right) = \frac{1}{\sqrt{2}}\]

\[ \Rightarrow \frac{\phi}{2} = \frac{\pi}{4}\]

\[ \Rightarrow \phi = \frac{\pi}{2}\]

Corrosponding path difference, \[∆ x = \frac{\lambda}{4}\]

\[ \Rightarrow y = \frac{∆ xD}{d} = \frac{\lambda D}{4d}\]

(b) When the intensity is one-fourth of the maximum:-

\[\frac{I}{I_\max} = \frac{1}{4}\]

\[ \Rightarrow 4 a^2  \cos^2 \left( \frac{\phi}{2} \right) = \frac{1}{4}\]

\[ \Rightarrow  \cos^2   \left( \frac{\phi}{2} \right) = \frac{1}{4}\]

\[ \Rightarrow \cos\left( \frac{\phi}{2} \right) = \frac{1}{2}\]

\[ \Rightarrow \frac{\phi}{2} = \frac{\pi}{3}\]

So, corrosponding path difference, \[∆ x = \frac{\lambda}{3}\]

and position, \[y = \frac{∆ xD}{d} = \frac{\lambda D}{3d}.\]