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Question: The ratio of maximum to minimum intensity due to superposition of two waves is \(\dfrac{{49}}{9}\) ....

The ratio of maximum to minimum intensity due to superposition of two waves is 499\dfrac{{49}}{9} . Then the ratio of intensity of component waves is:
A. 254{\text{A}}{\text{. }}\dfrac{{25}}{4}
B. 1625{\text{B}}{\text{. }}\dfrac{{16}}{{25}}
C. 449{\text{C}}{\text{. }}\dfrac{4}{{49}}
D. 949{\text{D}}{\text{. }}\dfrac{9}{{49}}

Explanation

Solution

- Hint – The ratio of maximum to minimum intensity due to superposition is given by Imax⁡Imin⁡=(a+b)2(a−b)2\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = \dfrac{{{{(a + b)}^2}}}{{{{(a - b)}^2}}} , where a and b are the amplitudes of the two waves.
Formula used- Imax⁡Imin⁡=(a+b)2(a−b)2\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = \dfrac{{{{(a + b)}^2}}}{{{{(a - b)}^2}}} , I1I2=(a)2(b)2\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{{{(a)}^2}}}{{{{(b)}^2}}}

Complete step-by-step solution -

We have been given the question that the ratio of maximum to minimum intensity due to superposition of two waves is 499\dfrac{{49}}{9} .
So, as we know that the ratio of maximum to minimum intensity due to superposition is given by the formula-
Imax⁡Imin⁡=(a+b)2(a−b)2\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = \dfrac{{{{(a + b)}^2}}}{{{{(a - b)}^2}}}
Here, a and b are the amplitudes of the two waves.
We have been given this ratio is equal to 499\dfrac{{49}}{9} , so putting this in the above formula we get-
Imax⁡Imin⁡=(a+b)2(a−b)2=499\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = \dfrac{{{{(a + b)}^2}}}{{{{(a - b)}^2}}} = \dfrac{{49}}{9}
We can write the above equation as-
(a+b)2(a−b)2=499\dfrac{{{{(a + b)}^2}}}{{{{(a - b)}^2}}} = \dfrac{{49}}{9}
Taking square root both sides we get-
(a+b)(a−b)=73\dfrac{{(a + b)}}{{(a - b)}} = \dfrac{7}{3}
on cross multiplying we get-
3(a+b)=7(a−b)3(a + b) = 7(a - b)
Solving further we get-
3a+3b=7a−7b ⇒4a=10b ⇒ab=104=52  3a + 3b = 7a - 7b \\\ \Rightarrow 4a = 10b \\\ \Rightarrow \dfrac{a}{b} = \dfrac{{10}}{4} = \dfrac{5}{2} \\\
Also, now using the formula that the ratio of intensity of component waves is given by-
I1I2=(a)2(b)2\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{{{(a)}^2}}}{{{{(b)}^2}}}
Now we have the value of ab=52\dfrac{a}{b} = \dfrac{5}{2} putting this in above formula, we get-
I1I2=(a)2(b)2=(ab)2=(52)2=254\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{{{(a)}^2}}}{{{{(b)}^2}}} = {\left( {\dfrac{a}{b}} \right)^2} = {\left( {\dfrac{5}{2}} \right)^2} = \dfrac{{25}}{4}
Therefore, the ratio of intensity of component waves is 25: 4.
Hence, the correct option is A.

Note – The principle of superposition may be applied to waves whenever two (or more) waves travelling through the same medium at the same time. Here, in the question as it is given that the ratio of maximum to minimum intensity due to superposition is given as 49/9, so using the formula as mentioned in the question, first find the ratio of a and b, which are the amplitudes and then put the value a/b in the formula I1I2=(a)2(b)2\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{{{(a)}^2}}}{{{{(b)}^2}}} , to find the ratio of intensity of component waves.