Question

Two coherent sources, whose intensity ratio is 81 : 1 produce interference fringes. Calculate the ratio of intensity of maxima and minima in the fringe system.
1. 16/25
2. 25/16
3. 9/1
4. 3/1

Answer 1

Correct Answer - Option 2 : 25/16

Concept:

Intensity of light ∝  width of the slit (W)

\(\frac{{{W_1}}}{{{W_2}}} = \frac{{{I_1}}}{{{I_2}}}\)

Intensity ∝ square of amplitude

\(\frac{{{W_1}}}{{{W_2}}} = \frac{{{I_1}}}{{{I_2}}} = \frac{{a_1^2}}{{a_2^2}}\)

\( \frac{{{I_{max}}}}{{{I_{min}}}} = \frac{{{{\left( {{a_1} + {a_2}} \right)}^2}}}{{{{\left( {{a_1} - {a_2}} \right)}^2}}} \)

Calculation:

\(\frac{{{I_1}}}{{{I_2}}} = \frac{{81}}{1}\)

\(\begin{array}{l} \frac{{{I_1}}}{{{I_2}}} = \frac{{a_1^2}}{{a_2^2}} = \frac{{81}}{1}\Rightarrow \frac{{{a_1}}}{{{a_2}}} = \frac{9}{1} \end{array}\)

\(\begin{array}{l} \frac{{{I_{max}}}}{{{I_{min}}}} = \frac{{{{\left( {{a_1} + {a_2}} \right)}^2}}}{{{{\left( {{a_1} - {a_2}} \right)}^2}}} = \frac{{{{\left( {\frac{{{a_1}}}{{{a_2}}} + 1} \right)}^2}}}{{{{\left( {\frac{{{a_1}}}{{{a_2}}} - 1} \right)}^2}}} = \frac{{{{\left( {\frac{9}{1} + 1} \right)}^2}}}{{{{\left( {\frac{9}{1} - 1} \right)}^2}}} = {\left( {\frac{10}{8}} \right)^2} = \frac{25}{16} \end{array}\)