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.