Answer
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Hint: The ratio of the intensities can be determined by using the given information of the average of the intensities. The average of the shows the maximum and minimum intensities, the maximum and minimum intensities shows the intensities of the two sources.
Complete step by step solution
Given that,
The interference pattern obtained with the help of two coherent sources is $5\% $ of the average intensity.
Now,
The average of the two intensities is given by,
${I_{avg}} = \dfrac{{{I_{\max }} + {I_{\min }}}}{2}$
Now, the maximum and minimum intensities of the two sources are given by,
For maximum,
${I_{\max }} = {\left( {\sqrt {{I_a}} + \sqrt {{I_b}} } \right)^2}$
For minimum,
${I_{\min }} = {\left( {\sqrt {{I_a}} - \sqrt {{I_b}} } \right)^2}$
Where, ${I_a}$ and ${I_b}$ are the intensities of the two sources.
Now, by the given data, the intensity variation of the interference pattern is $5\% $ of the average intensity, then
${I_{\max }} - {I_{\min }} = 5\% \times {I_{avg}}$
On further steps in the above equation, then
${I_{\max }} - {I_{\min }} = \dfrac{5}{{100}} \times {I_{avg}}$
Now substituting the average of intensities formula in the above equation, then
${I_{\max }} - {I_{\min }} = \dfrac{5}{{100}} \times \dfrac{{{I_{\max }} + {I_{\min }}}}{2}$
On dividing the terms in the above equation, then
${I_{\max }} - {I_{\min }} = \dfrac{1}{{20}} \times \dfrac{{{I_{\max }} + {I_{\min }}}}{2}$
On multiplying the terms in the above equation, then
${I_{\max }} - {I_{\min }} = \dfrac{{{I_{\max }} + {I_{\min }}}}{{40}}$
By rearranging the terms in the above equation, then
$40\left( {{I_{\max }} - {I_{\min }}} \right) = {I_{\max }} + {I_{\min }}$
By multiplying the terms in the LHS, then
$40{I_{\max }} - 40{I_{\min }} = {I_{\max }} + {I_{\min }}$
By rearranging the terms in the above equation, then
$40{I_{\max }} - 40{I_{\min }} - {I_{\max }} - {I_{\min }} = 0$
By subtracting the terms in the above equation, then
$39{I_{\max }} - 41{I_{\min }} = 0$
By rearranging the terms in the above equation, then
$39{I_{\max }} = 41{I_{\min }}$
By rearranging the terms in the above equation, then
$\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = \dfrac{{41}}{{39}}$
By substituting the formula of ${I_{\max }}$ and ${I_{\min }}$ in the above equation, then
$\dfrac{{{{\left( {\sqrt {{I_a}} + \sqrt {{I_b}} } \right)}^2}}}{{{{\left( {\sqrt {{I_a}} - \sqrt {{I_b}} } \right)}^2}}} = \dfrac{{41}}{{39}}$
The above equation is also written as,
${\left( {\dfrac{{\sqrt {{I_a}} + \sqrt {{I_b}} }}{{\sqrt {{I_a}} - \sqrt {{I_b}} }}} \right)^2} = \dfrac{{41}}{{39}}$
By taking square root on both sides, then the above equation is written as,
$\dfrac{{\sqrt {{I_a}} + \sqrt {{I_b}} }}{{\sqrt {{I_a}} - \sqrt {{I_b}} }} = \sqrt {\dfrac{{41}}{{39}}} $
The above equation is also written as,
$\dfrac{{\sqrt {{I_a}} + \sqrt {{I_b}} }}{{\sqrt {{I_a}} - \sqrt {{I_b}} }} = \dfrac{{\sqrt {41} }}{{\sqrt {39} }}$
By cross multiplying the terms in the above equation, then
$\sqrt {39} \left( {\sqrt {{I_a}} + \sqrt {{I_b}} } \right) = \sqrt {41} \left( {\sqrt {{I_a}} - \sqrt {{I_b}} } \right)$
By multiplying the terms in the above equation, then
$\sqrt {39} \sqrt {{I_a}} + \sqrt {39} \sqrt {{I_b}} = \sqrt {41} \sqrt {{I_a}} - \sqrt {41} \sqrt {{I_b}} $
By rearranging the terms in the above equation, then
$\sqrt {41} \sqrt {{I_b}} + \sqrt {39} \sqrt {{I_b}} = \sqrt {41} \sqrt {{I_a}} - \sqrt {39} \sqrt {{I_a}} $
By taking the terms as common in both sides, then
$\sqrt {{I_b}} \left( {\sqrt {41} + \sqrt {39} } \right) = \sqrt {{I_a}} \left( {\sqrt {41} - \sqrt {39} } \right)$
By rearranging the terms in the above equation, then
$\dfrac{{\left( {\sqrt {41} + \sqrt {39} } \right)}}{{\left( {\sqrt {41} - \sqrt {39} } \right)}} = \dfrac{{\sqrt {{I_a}} }}{{\sqrt {{I_b}} }}$
The above equation is also written as,
$\sqrt {\dfrac{{{I_a}}}{{{I_b}}}} = \dfrac{{\sqrt {41} + \sqrt {39} }}{{\sqrt {41} - \sqrt {39} }}$
By taking square on both side in the above equation, then
$\dfrac{{{I_a}}}{{{I_b}}} = {\left( {\dfrac{{\sqrt {41} + \sqrt {39} }}{{\sqrt {41} - \sqrt {39} }}} \right)^2}$
Thus, the above equation shows the ratio of the intensities of two sources.
Note: The two sources of the intensities are $a$ and $b$, so the intensities of the two sources are ${I_a}$ and ${I_b}$. The maximum intensity of the two intensities are equal to the sum of the square root of the individual intensities. The minimum intensity of the two intensities are equal to the difference of the square root of the individual intensities.
Complete step by step solution
Given that,
The interference pattern obtained with the help of two coherent sources is $5\% $ of the average intensity.
Now,
The average of the two intensities is given by,
${I_{avg}} = \dfrac{{{I_{\max }} + {I_{\min }}}}{2}$
Now, the maximum and minimum intensities of the two sources are given by,
For maximum,
${I_{\max }} = {\left( {\sqrt {{I_a}} + \sqrt {{I_b}} } \right)^2}$
For minimum,
${I_{\min }} = {\left( {\sqrt {{I_a}} - \sqrt {{I_b}} } \right)^2}$
Where, ${I_a}$ and ${I_b}$ are the intensities of the two sources.
Now, by the given data, the intensity variation of the interference pattern is $5\% $ of the average intensity, then
${I_{\max }} - {I_{\min }} = 5\% \times {I_{avg}}$
On further steps in the above equation, then
${I_{\max }} - {I_{\min }} = \dfrac{5}{{100}} \times {I_{avg}}$
Now substituting the average of intensities formula in the above equation, then
${I_{\max }} - {I_{\min }} = \dfrac{5}{{100}} \times \dfrac{{{I_{\max }} + {I_{\min }}}}{2}$
On dividing the terms in the above equation, then
${I_{\max }} - {I_{\min }} = \dfrac{1}{{20}} \times \dfrac{{{I_{\max }} + {I_{\min }}}}{2}$
On multiplying the terms in the above equation, then
${I_{\max }} - {I_{\min }} = \dfrac{{{I_{\max }} + {I_{\min }}}}{{40}}$
By rearranging the terms in the above equation, then
$40\left( {{I_{\max }} - {I_{\min }}} \right) = {I_{\max }} + {I_{\min }}$
By multiplying the terms in the LHS, then
$40{I_{\max }} - 40{I_{\min }} = {I_{\max }} + {I_{\min }}$
By rearranging the terms in the above equation, then
$40{I_{\max }} - 40{I_{\min }} - {I_{\max }} - {I_{\min }} = 0$
By subtracting the terms in the above equation, then
$39{I_{\max }} - 41{I_{\min }} = 0$
By rearranging the terms in the above equation, then
$39{I_{\max }} = 41{I_{\min }}$
By rearranging the terms in the above equation, then
$\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = \dfrac{{41}}{{39}}$
By substituting the formula of ${I_{\max }}$ and ${I_{\min }}$ in the above equation, then
$\dfrac{{{{\left( {\sqrt {{I_a}} + \sqrt {{I_b}} } \right)}^2}}}{{{{\left( {\sqrt {{I_a}} - \sqrt {{I_b}} } \right)}^2}}} = \dfrac{{41}}{{39}}$
The above equation is also written as,
${\left( {\dfrac{{\sqrt {{I_a}} + \sqrt {{I_b}} }}{{\sqrt {{I_a}} - \sqrt {{I_b}} }}} \right)^2} = \dfrac{{41}}{{39}}$
By taking square root on both sides, then the above equation is written as,
$\dfrac{{\sqrt {{I_a}} + \sqrt {{I_b}} }}{{\sqrt {{I_a}} - \sqrt {{I_b}} }} = \sqrt {\dfrac{{41}}{{39}}} $
The above equation is also written as,
$\dfrac{{\sqrt {{I_a}} + \sqrt {{I_b}} }}{{\sqrt {{I_a}} - \sqrt {{I_b}} }} = \dfrac{{\sqrt {41} }}{{\sqrt {39} }}$
By cross multiplying the terms in the above equation, then
$\sqrt {39} \left( {\sqrt {{I_a}} + \sqrt {{I_b}} } \right) = \sqrt {41} \left( {\sqrt {{I_a}} - \sqrt {{I_b}} } \right)$
By multiplying the terms in the above equation, then
$\sqrt {39} \sqrt {{I_a}} + \sqrt {39} \sqrt {{I_b}} = \sqrt {41} \sqrt {{I_a}} - \sqrt {41} \sqrt {{I_b}} $
By rearranging the terms in the above equation, then
$\sqrt {41} \sqrt {{I_b}} + \sqrt {39} \sqrt {{I_b}} = \sqrt {41} \sqrt {{I_a}} - \sqrt {39} \sqrt {{I_a}} $
By taking the terms as common in both sides, then
$\sqrt {{I_b}} \left( {\sqrt {41} + \sqrt {39} } \right) = \sqrt {{I_a}} \left( {\sqrt {41} - \sqrt {39} } \right)$
By rearranging the terms in the above equation, then
$\dfrac{{\left( {\sqrt {41} + \sqrt {39} } \right)}}{{\left( {\sqrt {41} - \sqrt {39} } \right)}} = \dfrac{{\sqrt {{I_a}} }}{{\sqrt {{I_b}} }}$
The above equation is also written as,
$\sqrt {\dfrac{{{I_a}}}{{{I_b}}}} = \dfrac{{\sqrt {41} + \sqrt {39} }}{{\sqrt {41} - \sqrt {39} }}$
By taking square on both side in the above equation, then
$\dfrac{{{I_a}}}{{{I_b}}} = {\left( {\dfrac{{\sqrt {41} + \sqrt {39} }}{{\sqrt {41} - \sqrt {39} }}} \right)^2}$
Thus, the above equation shows the ratio of the intensities of two sources.
Note: The two sources of the intensities are $a$ and $b$, so the intensities of the two sources are ${I_a}$ and ${I_b}$. The maximum intensity of the two intensities are equal to the sum of the square root of the individual intensities. The minimum intensity of the two intensities are equal to the difference of the square root of the individual intensities.
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