Question

Ratio of intensities of two waves is $9:1$ . If these two are superimposed, what is the ratio of maximum and minimum intensities:
A. $3:1$
B. $4:9$
C. $4:1$
D. $1:9$

Answer 1

Hint:In order to answer this question, first we will assume the intensities be ${I_1}\,and\,{I_2}$ and then we will write the ratio between them in terms of their amplitudes. And then we can find the ratio of maximum and minimum intensities with the help of their amplitudes.

Complete step by step answer:
Given that- Ratio of intensities of two waves is $9:1$.Let the two intensities be ${I_1}\,and\,{I_2}$.So, the ratio between both the intensities are as follow:
$\dfrac{{{I_1}}}{{{I_2}}} = {(\dfrac{{{a_1}}}{{{a_2}}})^2} = \dfrac{9}{1}$
where, ${a_1}\,and\,{a_2}\,$ are the amplitudes of ${I_1}\,and\,{I_2}$ respectively.
$ \Rightarrow {a_1} = 3{a_2}$ ………..eq(i)

Now, the ratio of the Maximum and minimum intensities are as follows:-
$\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = \dfrac{{{{({a_1} + {a_2})}^2}}}{{{{({a_1} - {a_2})}^2}}}$
Now we will use equation(i) in the above equation:-
$ \Rightarrow \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{{{(3{a_2} + {a_2})}^2}}}{{{{(3{a_2} - {a_2})}^2}}} \\
 \Rightarrow \dfrac{{{I_1}}}{{{I_2}}}= {(\dfrac{{4{a_2}}}{{2{a_2}}})^2} \\
 \therefore \dfrac{{{I_1}}}{{{I_2}}}= \dfrac{4}{1}$
Therefore, the ratio of maximum and minimum intensities is $4:1$.

Hence, the correct option is C.

Note:The intensity of a wave at one distance and the same wave's intensity at a different distance The power per unit area of a wave is characterised as its intensity, and as area varies with distance from the source, intensity is strongly related to power through distance.