Correct Answer - Option 3 : 64 ∶ 4
Concept:
The intensity of the light ∝ width of the slit (W)
\(\frac{{{W_1}}}{{{W_2}}} = \frac{{{I_1}}}{{{I_2}}}\)
Intensity ∝ square of the 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:
Given that, I1 = 25, I2 = 9
\(\frac{{{I_1}}}{{{I_2}}} = \frac{{25}}{9}\)
\(\begin{array}{l} \frac{{{I_1}}}{{{I_2}}} = \frac{{a_1^2}}{{a_2^2}} = \frac{{25}}{9}\Rightarrow \frac{{{a_1}}}{{{a_2}}} = \frac{5}{3} \end{array}\)
\(\begin{array}{l} \frac{{{I_{max}}}}{{{I_{min}}}} = \frac{{{{\left( {{a_1} + {a_2}} \right)}^2}}}{{{{\left( {{a_1} - {a_2}} \right)}^2}}} = \frac{{{{\left( {\frac{{{a_1}}}{{{a_2}}} + 1} \right)}^2}}}{{{{\left( {\frac{{{a_1}}}{{{a_2}}} - 1} \right)}^2}}} = \frac{{{{\left( {\frac{5}{3} + 1} \right)}^2}}}{{{{\left( {\frac{5}{3} - 1} \right)}^2}}} = {\left( {\frac{8}{2}} \right)^2} = \frac{64}{4} \end{array}\)