Question

The correct formula for fringe visibility is
A. $V = \dfrac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}}$
B. $V = \dfrac{{{I_{\max }} + {I_{\min }}}}{{{I_{\max }} - {I_{\min }}}}$
C. \[V = \dfrac{{{I_{\max }}}}{{{I_{\min }}}}\]
D. \[V = \dfrac{{{I_{\min }}}}{{{I_{\max }}}}\]

Answer 1

Hint: You can start by briefly explaining interference. Also include the concept of constructive (maximum intensity) and destructive (minimum intensity) interference. Then write the equation for fringe visibility, i.e. $V = \dfrac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}}$ .

Complete answer:
In physics, interference is a very important phenomenon. Interference involves the superimposition of two waves which form a resultant with amplitude more, less, or the same as the two waves involved. For interference, it is very important that the two waves either come from the same source or have the same or nearly the same frequency. Interference occurs in all types of waves including but not limited to acoustic, surface water waves, gravity waves, matter water, and light waves.
All this discussion was to make it easier to understand the concept of constructive and destructive interference.
Constructive interference occurs only when the path difference between the two waves is an integral multiple of the wavelength of the wave. Here the intensity of light becomes maximum which is called \[{I_{\max }}\] .
Destructive interference occurs only when the path difference between the two waves is an odd integral multiple of half wavelength of the wave. Here the intensity of light becomes minimum which is called \[{I_{\min }}\] .
The correct formula for fringe visibility is
$V = \dfrac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}}$
Fringe visibility describes how good the interference pattern is formed by the superimposition of two waves is. In simple words, in the interference pattern, you have bright parts (maximum intensity) and some parts that are dim in comparison (minimum intensity). Fringe visibility refers to how good we can distinguish between the bright and dim parts of the interference parts.

So, the correct answer is “Option A”.

Note:
As discussed earlier to produce constructive and destructive interference, we need two waves from the same source or with a constant phase difference. In labs, white light is not usually used to demonstrate interference, normally we use sodium lamps as a single light source for the two waves. If we do not use waves from the same source then the interference pattern will not be proper.