Question

In an interference of light derived from two slit apertures, if at some point on the screen, yellow light has a path difference of $\frac{3\lambda}{2}$, then the fringe at that point will be :

• A. yellow in colour
• B. white in colour
• C. dark
• D. bright

Answer 1

Answer: (C)

dark

Answer 2

The problem describes an interference pattern formed by light from two slit apertures. We are given the path difference at a certain point on the screen for yellow light and need to determine the nature of the fringe at that point.

In Young's Double Slit Experiment (YDSE), the conditions for constructive and destructive interference are:

1. For Constructive Interference (Bright Fringe):

The path difference ($\Delta x$) must be an integral multiple of the wavelength ($\lambda$).

$\Delta x = n\lambda$, where $n = 0, 1, 2, \ldots$

2. For Destructive Interference (Dark Fringe):

The path difference ($\Delta x$) must be an odd multiple of half the wavelength ($\lambda/2$).

$\Delta x = (n + \frac{1}{2})\lambda$ or $(2n + 1)\frac{\lambda}{2}$, where $n = 0, 1, 2, \ldots$

Given in the problem, the path difference for yellow light is $\Delta x = \frac{3\lambda}{2}$.

Let's compare this given path difference with the conditions for bright and dark fringes:

• Check for Bright Fringe:

If $\Delta x = n\lambda$, then $\frac{3\lambda}{2} = n\lambda \implies n = \frac{3}{2}$.

Since $n$ must be an integer for a bright fringe, this condition is not met.

• Check for Dark Fringe:

If $\Delta x = (n + \frac{1}{2})\lambda$, then $\frac{3\lambda}{2} = (n + \frac{1}{2})\lambda$.

Dividing both sides by $\lambda$:

$\frac{3}{2} = n + \frac{1}{2}$

$n = \frac{3}{2} - \frac{1}{2}$

$n = \frac{2}{2}$

$n = 1$.

Since $n=1$ is an integer, this condition for destructive interference is met.

Therefore, the fringe at that point will be dark.