Question

In a Young's double slit experiment, \[\lambda = 500\text{ nm, d = 1.0 mm and D = 1.0 m.}\] Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.

Answer 1

Given:-

Separation between the two slits,

\[d = 1 mm = {10}^{- 3} m\]

Wavelength of the light,

\[\lambda = 500 nm = 5 \times {10}^{- 7} m\]

Distance of the screen,

\[D = 1 m\]

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, \[∆ x = \frac{1}{4}\]

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

\[\Rightarrow y = \frac{5 \times {10}^{- 7} \times 1}{4 \times {10}^{- 3}}\]

\[ = 1 . 25 \times {10}^{- 4} m\]

∴ The required minimum distance from the central maximum is \[1 . 25 \times {10}^{- 4} m.\]