Question

In Young’s double-slit experiment using monochromatic light of wavelength \(λ\), the intensity of light at a point on the screen where path difference is \(λ\), is k units. What is the intensity of light at a point where path difference is \(\frac{λ }{3}\)?

Answer 1

The correct answer is: \(\frac{Κ}{4}\)
Let \(I_1\) and \(I_2\) be the intensity of the two light waves. Their resultant intensities can be obtained as: 
\(I' = I_1+I_2+2\sqrt{I_1I_2}cos\phi\)
Where,
\(\phi\) = Phase difference between the two waves
For monochromatic light waves,
\(I_1 = I_2\)
\(∴ I' = I_1+I_2+2\sqrt{I_1I_1} cos\phi\)
\(= 2I_1+2I_1 cos\phi\)
Phase difference =\( \frac{2π }{ λ} ×\) path difference
Since path difference \(= λ, \)
Phase difference, \(\phi= 2π\)
\(∴ I' = 2I_1+2I_1 = 4I_1\)
Given
\(I' = K\)
\(∴ I_1 = \frac{k}{4} \)                        ...(1)
When path difference \(= \frac{λ}{3},\)
Phase difference, \(\phi = \frac{2π}{3}\)
Hence, resultant intensity, \(I'_R = I_1+I_1+2\sqrt{I_1I_1}cos \frac{2π}{3}\)
\(= 2I_1+2I_1(\frac{-1}{2})=I_1\)
Using equation (1), we can write:
\(I_R = I_1 = \frac{k}{4}\)
Hence, the intensity of light at a point where the path difference is \(\frac{λ}{3}\) is \(\frac{Κ}{4}\) units.