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

The ratio of maximum to minimum intensity due to superposition of two waves is 499\dfrac{49}{9}. Then the ratio of the intensity of component waves is
(a) 254\dfrac{25}{4}
(b) 1625\dfrac{16}{25}
(c) 449\dfrac{4}{49}
(d) 949\dfrac{9}{49}

Explanation

Solution

Hint: To find the ratio of maximum and minimum intensity due to superposition of two waves can be given by ratio of addition and subtraction of amplitudes, which can be given mathematically as, I⁡maxI⁡min⁡=(a+b)2(a−b)2\dfrac{{{\operatorname{I}}_{\text{max}}}}{{{\operatorname{I}}_{\min }}}=\dfrac{{{\left( a+b \right)}^{2}}}{{{\left( a-b \right)}^{2}}}, form these we will find the values of amplitudes and from that we will find the ratio of intensities by using the formula, I⁡1I⁡2=(a)2(b)2\dfrac{{{\operatorname{I}}_{1}}}{{{\operatorname{I}}_{2}}}=\dfrac{{{\left( a \right)}^{2}}}{{{\left( b \right)}^{2}}}.

Formula used: I⁡maxI⁡min⁡=(a+b)2(a−b)2\dfrac{{{\operatorname{I}}_{\text{max}}}}{{{\operatorname{I}}_{\min }}}=\dfrac{{{\left( a+b \right)}^{2}}}{{{\left( a-b \right)}^{2}}}, I⁡1I⁡2=(a)2(b)2\dfrac{{{\operatorname{I}}_{1}}}{{{\operatorname{I}}_{2}}}=\dfrac{{{\left( a \right)}^{2}}}{{{\left( b \right)}^{2}}}

Complete step by step answer:
We know that intensity of any wave is directly proportional to the square of its amplitude, which can be given as, I∝a2I\propto {{a}^{2}}.
Now, the maxima and minima of waves can be given by the formula,
I⁡maxI⁡min⁡=(a+b)2(a−b)2\dfrac{{{\operatorname{I}}_{\text{max}}}}{{{\operatorname{I}}_{\min }}}=\dfrac{{{\left( a+b \right)}^{2}}}{{{\left( a-b \right)}^{2}}} ………………..(i)
Where, a and b are amplitudes of waves.
Now, in question we are given that ratio of maximum and minimum intensity of superposition is 499\dfrac{49}{9} so, on substituting the value in expression (i) we will get,
I⁡maxI⁡min⁡=(a+b)2(a−b)2=499\dfrac{{{\operatorname{I}}_{\text{max}}}}{{{\operatorname{I}}_{\min }}}=\dfrac{{{\left( a+b \right)}^{2}}}{{{\left( a-b \right)}^{2}}}=\dfrac{49}{9}
⇒a+ba−b=499\Rightarrow \dfrac{a+b}{a-b}=\sqrt{\dfrac{49}{9}}
⇒a+ba−b=73⇒3(a+b)=7(a−b)\Rightarrow \dfrac{a+b}{a-b}=\dfrac{7}{3}\Rightarrow 3\left( a+b \right)=7\left( a-b \right)
⇒3a+3b=7a−7b\Rightarrow 3a+3b=7a-7b
⇒3b+7b=7a−3a\Rightarrow 3b+7b=7a-3a
⇒10b=4a⇒104=ab\Rightarrow 10b=4a\Rightarrow \dfrac{10}{4}=\dfrac{a}{b}
⇒ab=52\Rightarrow \dfrac{a}{b}=\dfrac{5}{2} ………………..(ii)
Now, we know that intensities can be given by the formula, I∝a2I\propto {{a}^{2}}, so, the ratio of intensities can be given by the formula,
⇒I1I2=a2b2\Rightarrow \dfrac{{{I}_{1}}}{{{I}_{2}}}=\dfrac{{{a}^{2}}}{{{b}^{2}}}
⇒I1I2=5222=254\Rightarrow \dfrac{{{I}_{1}}}{{{I}_{2}}}=\dfrac{{{5}^{2}}}{{{2}^{2}}}=\dfrac{25}{4}
Hence, the ratio of component intensities is 254\dfrac{25}{4}.
Thus, option (a) is correct.

Note: Interference is a phenomenon which takes place in young’s single slit and double slit experiment. Now, the point at which amplitudes of two waves add up is called maxima or the intensity is maximum and the point at which amplitudes of two waves cancel each other is called minima of the intensity is minimum.