An electromagnetic wave can be represented by `E = A sin (kx- omega t + phi)`, where E is electric field associated with wave, According this equation, for any value of `x, E` remains sinusoidal for `-oolt t lt oo`. Obviously this corresponds to an idealised situation because radiation from ordinary sources consists of finite size wavetrains. In general, electric field remains sinusoidal only for times of order `tau_(c)`' which is called coherence time. In simpler language it means that for times of order `tau_(c)'` `a` wave will have a definite phase. The finite value of coherence time could be due to many factors, for example if radiating atom undergoes collision with another atom then wave train undergoes an abrupt phase change or due to the fact that an atom responsible for emitting radiation has a finite life time in the energy level from which it drops to lower energy level, while radiating. Concept of coherence time can be easily understood using young's double slit experiment. Let interference patten is observed around point `P` at time `t`, due to superposition of waves emanting from `S_(1)` and `S_(2)` at times `t =(r_(1))/(c)` and `(r_(2))/(c)` respectively, where `r_(1)` and `r_(2) ` are the distances`S_(1) P & S_(2)P`. Obviously if `(r_(2)-r_(1))/(c) lt lt tau_(e),{"where" " "c = 3xx10^(8)m//s}` then, wavetrain arriving at point `P` from `S_(1)` & `S_(2)` will have a definite phase relationship and an interference pattern of good contranst will be obtained.

If coherence time is of order `10^(-10)` second and screen is placed at a very large distance from slits in the given figure, then:-

Options

A. For `d = 0.3 cm`, we will not get any fringers on screen (except central maxima)
B. For `d = 0.03 cm`, we will get fringes on a large part of screen
C. For `d = 3 cm` we will get fringes on entire screen.
D. We will get fringes on entire screen, irrespective of value of `d`.

Answer

B

Explanation

`d sin theta = (10^(-10))(3xx10^(8)) = 3xx10^(-2)m = 3cm`
`(A)` for `d = 0.3 cm`, we will get fringes on entire screen
`(B)` For `d = 0.03 cm`, we will get fringes on entire screen
`(C )` For `d = 3 cm, theta = 45^(@)`