There is a source emitting electromagnetic waves of wavelength 2m . One light ray reaches a point P in space direcly and the other ray reaches the same point P after travelling 1m extra distance and after getting reflected from a glass plate. Let I be the original intensity corresponding to both rays and glass plate reflects only 25% of incident energy. Resultant intensity at point P will be

Options

A. `3I//2`
B. `9I//2`
C. `9I//4`
D. `0`

Answer

C

Explanation

Let `I_1 and I_2` be the intensities corresponding to both rays, then we can write the following:
`I_1 = I`
`I_2 = I//4`, because 25% energy is refleced.
Path difference introduced is `Deltax = 1m` and thus correseponding phase difference can be written as
`delta. = (2pi//lambda) Delta x = (2 pi//2)(1) = pi`
There is an additional phase difference of `pi` due to reflection and hence net phase difference between the two rays becomes zero.
`delta = 0`
Resulant intensity can be written as follows:
`I = I_1 + I_2 + 2sqrt(I_1I_2) cos delta`
`implies I_("net") = I + I/4 + 2sqrt(I xx I/4) cos 0 = (9I)/4`