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} \\
 $