Question

Two waves of the same frequency and
constant phase difference have intensities in
the ratio 36 : 4. These waves are superposed
and interference fringe pattern is obtained.
Calculate the ratio of the maximum to
minimum intensity.

Answer 1

Answer on Question #77789, Physics / Optics

Question. Two waves of the same frequency and constant phase difference have intensities in the ratio $36:4$. These waves are superposed and interference fringe pattern is obtained. Calculate the ratio of the maximum to minimum intensity.

Given. $\frac{I_1}{I_2} = \frac{36}{4}$.

Find. $\frac{I_{max}}{I_{min}} - ?$

Solution.

So,

$$\frac{I_1}{I_2} = \frac{(a_1)^2}{(a_2)^2} = \left(\frac{a_1}{a_2}\right)^2 = \frac{36}{4} = \left(\frac{6}{2}\right)^2$$ $$\frac{a_1}{a_2} = \frac{6}{2}$$ $$\frac{I_{max}}{I_{min}} = \frac{(a_1 + a_2)^2}{(a_2 - a_2)^2} = \frac{(6 + 2)^2}{(6 - 2)^2} = \frac{64}{16} \text{ or } \frac{4}{1}$$

Answer. $\frac{I_{max}}{I_{min}} = \frac{64}{16}$.

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