Question

In an interference experiment the ratio of amplitudes of coherent waves is $ \dfrac{{{a_1}}}{{{a_2}}}{\text{ }} = {\text{ }}\dfrac{1}{3} $ . The ratio of maximum and minimum intensities of fringes will be:
A) 4
B) 2
C) 9
D) 18

Answer 1

Hint : The intensity of the fringes is proportional to the square of the amplitude of the coherent waves. The maximum intensity will correspond to the sum of the square of the individual coherent waves and the minimum intensity will correspond to the difference of the square of the individual reference.

Complete step by step answer
We’ve been given that the ratio of the amplitude of the coherent waves is
 $ \dfrac{{{a_1}}}{{{a_2}}}{\text{ }} = {\text{ }}\dfrac{1}{3} $
We know that the intensity of the fringes is proportional to the square of the amplitude of the coherent waves. So, the intensity of the individual intensities will be the ratio of the square of the amplitudes as:
 $ \dfrac{{{I_1}}}{{{I_2}}} = {\left( {\dfrac{{{a_1}}}{{{a_2}}}} \right)^2} = {\text{ }}\dfrac{1}{9} $
Let $ \dfrac{{{I_1}}}{{{I_2}}} = {x^2} $ , then $ {I_1} = {x^2} $ and $ {I_2} = 9{x^2} $ .
The maximum intensity will be
 $ {I_{max}} = \sqrt {{I_1}^2} + \sqrt {{I_2}^2} $
And the minimum intensity will be
 $ {I_{\min }} = \sqrt {{I_1}^2} - \sqrt {{I_2}^2} $
So, the ratio of the maximum and the minimum intensities will be the ratio of the two equations
 $ \dfrac{{{I_{max}}}}{{{I_{\min }}}} = {\left( {\dfrac{{\sqrt {{I_1}} + \sqrt {{I_2}} }}{{\sqrt {{I_1}} - \sqrt {{I_2}} }}} \right)^2} $
Substituting the values of $ {I_1} = {x^2} $ and $ {I_2} = 9{x^2} $ , we get
 $ \dfrac{{{I_{max}}}}{{{I_{\min }}}} = \left( {\dfrac{{\sqrt {{x^2}} + \sqrt {\left( {9{x^2}} \right)} }}{{\sqrt {{x^2}} - \sqrt {\left( {9{x^2}} \right)} }}} \right) $
Hence, we can simplify the above expression by taking out the $ {x^2} $ term common form the numerator and the denominator as
 $ \dfrac{{{I_{max}}}}{{{I_{\min }}}} = {\left( {\dfrac{{1 + 3}}{{1 - 3}}} \right)^2} $
Which is simplified as
 $ \dfrac{{{I_{max}}}}{{{I_{\min }}}} = 4 $
Hence the ratios of the maximum and minimum amplitude of the fringes will be $ \dfrac{{{I_{max}}}}{{{I_{\min }}}} = 4 $ which corresponds to option (A).

Note
We must be careful not to take the ratio of the amplitudes to find the ratio of the intensity. This is because the intensity is proportional to the square of the amplitude of the coherent waves and not the amplitude directly. The sources, even though have different amplitudes must be coherent to form an interference pattern that is satisfied in the question.