Question

If the amplitude ratio of two sources producing interference is $3:{\text{ }}5$, the ratio of intensities at maxima and minima is

$$A\,\,\,\,\,25:16 \\\ B\,\,\,\,\,5:3 \\\ C\,\,\,\,\,16:1 \\\ D\,\,\,\,25:9 \\\$$

Answer 1

Hint Using the amplitude ratio find the individual amplitudes and now substitute them in the intensity relation $\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = {\left( {\dfrac{{{a_1} + {a_2}}}{{{a_1} - {a_2}}}} \right)^2}$ to get the ratio in terms of, ${I_{max}}$ and ${I_{min}}$.

Complete step-by-step solution
In interference redistribution of energy takes place in the form of maxima and minima.
Ratio of maximum intensity and minimum intensity is given by
$\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = {\left( {\dfrac{{{a_1} + {a_2}}}{{{a_1} - {a_2}}}} \right)^2}$
Where, Imax and Imin are intensities at maxima and minima respectively.
$a_1$ and $a_2$ are the amplitude of the two sources.
Given that $${a_1}:{\text{ }}{a_2} = {\text{ }}3:{\text{ }}5$$$ \Rightarrow {a_1} = 3x;{a_2} = 5x$Substitute in the expression to get the intensity ratio$
\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = {\left( {\dfrac{{3x + 5x}}{{3x - 5x}}} \right)^2} \\
\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = {\left( {\dfrac{{8x}}{{ - 2x}}} \right)^2} \\
\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = {\left( {\dfrac{{ - 4}}{1}} \right)^2} \\
\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = \dfrac{{16}}{1} \\
$

Hence the ratio of intensities at maxima and minima is 16: 1 and the correct option is C.

Note Some relations regarding Imax and Imin with amplitudes $a_1$ and $a_2$
$\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} = \dfrac{{{a_1}}}{{{a_2}}}\left( {\dfrac{{\sqrt {\dfrac{{{I_{\max }}}}{{{I_{\min }}}}} + 1}}{{\sqrt {\dfrac{{{I_{\max }}}}{{{I_{\min }}}}} - 1}}} \right)$
${I_{avg}} = \dfrac{{{I_{\max }} + {I_{\min }}}}{2} = {I_1} + {I_2} = a_1^2 + a_2^2$
In constructive interference amplitude and intensity at the point of observation:
${A_{\max }} = {a_1} + {a_2} \\\ {I_{\max }} = {\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2} \\\$
In destructive interference amplitude and intensity at the point of observation:
${A_{\min }} = {a_1} - {a_2} \\\ {I_{\min }} = {\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2} \\\$