A source emits electromagnetic waves of wavelength `3 m`. One beam reaches the observer directly and other after reflection from a watersurface, travelling `1.5 m` extra distance and with intensity reduced to `1//4` as compared to intensity due to the direct beam alone. The resultant intensity will be :

Options

A. `(1//4) rad`
B. `(3//4) rad`
C. `(5//4) rad`
D. `(9//4) rad`

Answer

D

Explanation

As `Delta x = 1.5 m = (3)/(2)m = (lambda)/(2) :. Delta phi = pi` ,br. Reflection from water surface introduces a phase diff. of `pi`. Therefore, total phase diff. `= pi + pi = 2pi`.
`:.` Intensity obtained is maximum.
Now `I_(max) = I_(1) + I_(2) + 2sqrt(I_(1)I_(2)) cos 2pi`
`= I + (I)/(4) + 2sqrt(I xx I//4) = (9 I)/(4)`