The interference pattern is obtained with two coherent light sources of intensity ratio $n$. In the interference pattern, the ratio $\dfrac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}}$ will be:
(A) $\dfrac{{2\sqrt n }}{{{{\left( {n + 1} \right)}^2}}}$
(B) $\dfrac{{\sqrt n }}{{n + 1}}$
(C) $\dfrac{{2\sqrt n }}{{n + 1}}$
(D) $\dfrac{{\sqrt n }}{{{{\left( {n + 1} \right)}^2}}}$

Hint
In the interference pattern the ratio is determined by using the value of ${I_{\max }}$ and ${I_{\min }}$ , and also by using the given information of the ratio of the two coherent light source of intensities, then the ratio of the given form is determined.

Complete step by step answer
Given that, The interference pattern is obtained with two coherent light sources of intensity ratio $n$.
$\Rightarrow \dfrac{{{I_2}}}{{{I_1}}} = n$
By rearranging the terms in the above equation, then the above equation is written as,
$\Rightarrow {I_2} = n{I_1}$
Now, the intensity of the maximum is written as,
$\Rightarrow {I_{\max }} = {\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2}\,................\left( 1 \right)$
Now, the intensity of the minimum is written as,
$\Rightarrow {I_{\min }} = {\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2}\,..............\left( 2 \right)$
By substituting the given ratio of the two intensities in the equation (1), then the equation (1) is written as,
$\Rightarrow {I_{\max }} = {\left( {\sqrt {{I_1}} + \sqrt {n{I_1}} } \right)^2}$
By taking the common terms from the above equation, then the above equation is written as,
$\Rightarrow {I_{\max }} = \left[ {{{\left( {\sqrt {{I_1}} } \right)}^2}{{\left( {1 + \sqrt n } \right)}^2}} \right]$
By squaring the terms in the above equation, then the above equation is written as,
$\Rightarrow {I_{\max }} = {I_1}{\left( {1 + \sqrt n } \right)^2}$
From the mathematics ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$, substituting this formula in the above equation, then
$\Rightarrow {I_{\max }} = {I_1}\left( {1 + n + 2\sqrt n } \right)$
By substituting the given ratio of the two intensities in the equation (2), then the equation (2) is written as,
$\Rightarrow {I_{\min }} = {\left( {\sqrt {{I_1}} - \sqrt {n{I_1}} } \right)^2}$
By taking the common terms from the above equation, then the above equation is written as,
$\Rightarrow {I_{\min }} = \left[ {{{\left( {\sqrt {{I_1}} } \right)}^2}{{\left( {1 - \sqrt n } \right)}^2}} \right]$
By squaring the terms in the above equation, then the above equation is written as,
$\Rightarrow {I_{\max }} = {I_1}{\left( {1 - \sqrt n } \right)^2}$
From the mathematics ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$, substituting this formula in the above equation, then
$\Rightarrow {I_{\max }} = {I_1}\left( {1 + n - 2\sqrt n } \right)$
Then for the given ratio is, $\dfrac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}}$
By substituting the value of the ${I_{\max }}$ and ${I_{\min }}$ in the above equation, then
$\Rightarrow \dfrac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}} = \dfrac{{\left[ {{I_1}\left( {1 + n + 2\sqrt n } \right)} \right] - \left[ {{I_1}\left( {1 + n - 2\sqrt n } \right)} \right]}}{{\left[ {{I_1}\left( {1 + n + 2\sqrt n } \right)} \right] + \left[ {{I_1}\left( {1 + n - 2\sqrt n } \right)} \right]}}$
By taking the common term from the above equation, then
$\Rightarrow \dfrac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}} = \dfrac{{{I_1}\left[ {\left( {1 + n + 2\sqrt n } \right) - \left( {1 + n - 2\sqrt n } \right)} \right]}}{{{I_1}\left[ {\left( {1 + n + 2\sqrt n } \right) + \left( {1 + n - 2\sqrt n } \right)} \right]}}$
By cancelling the same terms from the above equation, then
$\Rightarrow \dfrac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}} = \dfrac{{\left[ {\left( {1 + n + 2\sqrt n } \right) - \left( {1 + n - 2\sqrt n } \right)} \right]}}{{\left[ {\left( {1 + n + 2\sqrt n } \right) + \left( {1 + n - 2\sqrt n } \right)} \right]}}$
By rearranging the terms in the above equation, then
$\Rightarrow \dfrac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}} = \dfrac{{1 + n + 2\sqrt n - 1 - n + 2\sqrt n }}{{1 + n + 2\sqrt n + 1 + n - 2\sqrt n }}$
On further simplification in the above equation, then
$\Rightarrow \dfrac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}} = \dfrac{{4\sqrt n }}{{2 + 2n}}$
By taking the common term from the above equation, then
$\Rightarrow \dfrac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}} = \dfrac{{4\sqrt n }}{{2\left( {1 + n} \right)}}$
On dividing the terms in the above equation, then
$\Rightarrow \dfrac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}} = \dfrac{{2\sqrt n }}{{\left( {1 + n} \right)}}$
Hence, the option (C) is the correct answer.

Note
The maximum intensity of the interference is the square of the sum of the square root of the individual intensities. The minimum intensity of the interference is the square of the difference of the square root of the individual intensities.