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Question: The correct formula for fringe visibility is: A. \(V = \dfrac{{{I_{\max }} - {I_{\min }}}}{{{I_...

The correct formula for fringe visibility is:

A. V=Imax⁡−Imin⁡Imax⁡+Imin⁡V = \dfrac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}}

B. V=Imax⁡+Imin⁡Imax⁡−Imin⁡V = \dfrac{{{I_{\max }} + {I_{\min }}}}{{{I_{\max }} - {I_{\min }}}}

C. V=Imax⁡Imin⁡V = \dfrac{{{I_{\max }}}}{{{I_{\min }}}}

D. V=Imin⁡Imax⁡V = \dfrac{{{I_{\min }}}}{{{I_{\max }}}}

Explanation

Solution

First let us see what fringe visibility is:
The interferometric visibility (otherwise called interference visibility and fringe visibility, or just visibility) evaluates the differentiation of interference in any framework which has wave-like properties, for example, optics, quantum mechanics, water waves, sound waves, or electrical signs.

Complete step by step answer:
In general, two or more are mixed and when the phase difference between them varies, the force or amplitude of the resulting wave oscillates, creating an interference pattern. The pattern can be visible all at once because the disparity between the phases changes as a function of space, as in a two slit experiment.
The ratio of the size or amplitude of these oscillations to the sum of the powers of each wave is known as visibility.
Interferometric visibility provides a realistic means to calculate the consistency of two waves with each other. The theoretical concept coherence is based on the degree of coherence, using the notion of correlation.
In linear optical interferometers, interference expresses itself as amplitude oscillations over time or space, also known as fringes. In these conditions interferometric visibility is known as fringe visibility. With this form of interference, the sum of the intensities of the two interference waves is equal to the total amplitude over a given time or space area. The formula for visibility is given by:
ν=AI‾\nu = \dfrac{A}{{\overline I }}

Amplitude is given by:
A=(Imax⁡−Imin⁡)2A = \dfrac{{\left( {{I_{\max }} - {I_{\min }}} \right)}}{2}

I‾=(Imax⁡−Imin⁡)2\overline I = \dfrac{{\left( {{I_{\max }} - {I_{\min }}} \right)}}{2}
So, now the visibility can be written as:
ν=Imax⁡−Imin⁡Imax⁡+Imin⁡\nu = \dfrac{{{I_{\max }} - {I_{\min }}}} {{{I_{\max }} + {I_{\min }}}}

Hence, option A is correct.

Note: Here we have to remember the formula of amplitude and intensity to get the formula for fringe width.
Also, we have to correctly remember the formula for visibility.