Two waves of intensity $l_{2} and I_{2}$ interfere. The maximum intensity produced to the minimum intensity is in the ratio a . Then, $I_{1} / I_{2}$ is equal to

Q. Two periodic waves of intensities I1 and I2 pass
through a region at the same time in the same
direction. the sum of the maximum and
minimum intensities is -

Q. Two beams of light of intensity

I_{1}

and

I_{2}

interfere to give an interference pattern. If the ratio of maximum intensity to that of minimum intensity is

\frac{25}{9}

, then ratio of

\frac{I_{1}}{I_{2}}

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Two light waves of intensities ′I′1 and ′I′2 having same frequency pass through same medium at a time in same direction and interfere. The sum of the minimum and maximum intensities is

The correct option is B 2(I1+I2)since the frequencies are same so the maximum intensity will be I=(√I1+√I2)2minimum intensity is i=(√I1−√I2)2addying both we get I+i=2×(I1+I2)so the answer is B.

Two light waves of intensities $^{'} I_{1}^{'}$ and $^{'} I_{2}^{'}$ having same frequency pass through same medium at a time in same direction and interfere. The sum of the minimum and maximum intensities is

The correct option is B $2 (I_{1} + I_{2}$ )
since the frequencies are same so
the maximum intensity will be $I = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
minimum intensity is $i = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
addying both we get $I + i = 2 \times (I_{1} + I_{2})$
so the answer is B.