1.

The interference pattern is obtained with twocoherent light sources of intensity ratio n. In theinterference pattern, the ratioImax -- /minwill be

Answer»

ANSWER:

Ratio of maximum and minimum intensity is GIVEN as

\frac{I_{max}}{I_{min}} = \frac{(\sqrt{n} + 1)^2}{(\sqrt{n} - 1)^2}

Explanation:

As we know that the ratio of two intensity of the given waves is

n = \frac{I_1}{I_2}

now we know that maximum intensity due to SUPERPOSITION of coherent waves is given by

I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2

similarly minimum FREQUENCY due to superposition of two coherent waves is given as

I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2

now we have to find the ratio of maximum and minimum intensity

so we have

\frac{I_{max}}{I_{min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2}

\frac{I_{max}}{I_{min}} = \frac{(\sqrt{I_1/I_2} + 1)^2}{(\sqrt{I_1/I_2} - 1)^2}

now we have

\frac{I_{max}}{I_{min}} = \frac{(\sqrt{n} + 1)^2}{(\sqrt{n} - 1)^2}

#Learn

Topic : Interference of coherent waves

brainly.in/question/7967009



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