A source emits electromagnetic waves of wavelength 3 m. One bema reaches the observer directly and other after reflection from a water surface, travelling 1.5 m extra distance and withintensity reduce to (1/4) as compared to intensity due to direct bema alone. The resultant intensity will be :

A source emits electromagnetic waves of wavelength 3 m. One bema reach

1.	A source emits electromagnetic waves of wavelength 3 m. One bema reaches the observer directly and other after reflection from a water surface, travelling 1.5 m extra distance and withintensity reduce to (1/4) as compared to intensity due to direct bema alone. The resultant intensity will be :
Answer» `(1/4)` fold `(3/4)` fold `(5/4)` fold `(9/4)` fold Solution :At reflection at DENSER medium, the phase difference is `pi` & path difference is `(LAMBDA)/(2)` `therefore` Total phase difference `= pi - pi = 0` HENCE, resultant amplitude is ` = sqrt I + sqrt(I)/(4) = 3/2 sqrtI` Then intensity = `(3/2 sqrtI)^(2) = 9/4 I`.

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A source emits electromagnetic waves of wavelength 3 m. One bema reaches the observer directly and other after reflection from a water surface, travelling 1.5 m extra distance and withintensity reduce to (1/4) as compared to intensity due to direct bema alone. The resultant intensity will be :

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