Question

Two coherent sources of intensity ratio $\beta$ interface. Then, the value of ${}^{({{I}_{Max}}-{{I}_{Min}})}/{}_{({{I}_{Max}}+{{I}_{Min}})}$ is
A. $\dfrac{1+\beta }{\sqrt{\beta }}$
B. $\sqrt{\dfrac{1+\beta }{\beta }}$
C. $\dfrac{1+\beta }{2\sqrt{\beta }}$
D. $\dfrac{2\sqrt{\beta }}{1+\beta }$

Answer 1

When two coherent sources superpose each other, then intensity in the region of superposition gets distributed becoming maxima at some points and minima at some points, this phenomenon is called Interface of light. In interference Intensity is directly proportional to square of amplitude.

Complete step by step answer:
Let us assume the intensity of two coherent sources is ${{I}_{1}}\And {{I}_{2}}$It is given in the question that the intensity ratio is$\beta$. So we can write Intensity ratio of two sources is$\dfrac{{{I}_{1}}}{{{I}_{2}}}=\beta$. Now, Let us assume that the amplitude of two coherent sources is${{A}_{1}}\And {{A}_{2}}$. Since we know the intensity of the wave is directly proportional to square to amplitude. So we can get, $\dfrac{{{I}_{1}}}{{{I}_{2}}}=\dfrac{{{A}_{1}}^{2}}{{{A}_{2}}^{2}}=\beta$
On Simplify the above result we get,
${{A}_{1}}^{2}=\beta {{A}_{2}}^{2}$ Square root on both sides, we get {{A}_{1}}=\sqrt{\beta }{{A}_{2}}$$$$(Equation_1) Let us assume the resultant intensity of two coherent sources is$I$. Resultant intensity can be expressed by mathematical relationship given as –
$I={{A}_{1}}^{2}+{{A}_{2}}^{2}+2{{A}_{1}}{{A}_{2}}Cos\phi$ This equation gives the total intensity at a point where the phase difference is φ. Here A1 and A2 are amplitudes of coherent sources. The total intensity relation contains a third term of cosine function on which whole value of intensity depends. That term is called interference term.
For maximum intensity$Cos\phi =1$. So phase difference between the two superposing waves is an even multiple of π or path difference is an integral multiple of wavelength. This is called Constructive Interference. So expression for maximum intensity can be written as- ${{I}_{\max }}={{({{A}_{1}}+{{A}_{2}})}^{2}}$
For minimum intensity $Cos\phi =-1$. So phase difference between the two superposing waves is an odd multiple of π or path difference is an odd multiple of half wavelength. This is called Destructive Interference. So expression for minimum intensity can be written as- ${{I}_{\min }}={{({{A}_{1}}-{{A}_{2}})}^{2}}$
$\because \dfrac{{{I}_{\max }}-{{I}_{\min }}}{{{I}_{\max }}+{{I}_{\min }}}=\dfrac{{{({{A}_{1}}+{{A}_{2}})}^{2}}-{{({{A}_{1}}-{{A}_{2}})}^{2}}}{{{({{A}_{1}}+{{A}_{2}})}^{2}}+{{({{A}_{1}}-{{A}_{2}})}^{2}}}$
\because \dfrac{{{I}_{\max }}-{{I}_{\min }}}{{{I}_{\max }}+{{I}_{\min }}}=\dfrac{4{{A}_{1}}{{A}_{2}}}{2({{A}_{1}}^{2}+{{A}_{2}}^{2})}$$$$=\dfrac{2{{A}_{1}}{{A}_{2}}}{({{A}_{1}}^{2}+{{A}_{2}}^{2})}
Put value of A1 from equation A in this equation as and on simplifying ,
We get,
$=\dfrac{2(\sqrt{\beta }{{A}_{2}}){{A}_{2}}}{\beta {{A}_{2}}^{2}+{{A}_{2}}^{2}}=\dfrac{2\sqrt{\beta }}{1+\beta }$ So, the correct answer is “Option D”.

Note: If cosine function term varies with time assuming both positive and negative values ,then the average value of cosine function will be zero .Then the interference term averages to zero, then there will be same intensity at every point on observation screen then the two sources in that case are incoherent.