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An electromagnetic wave of wavelength la...


An electromagnetic wave of wavelength `lamda_(0)` (in vacuum) passes from P towards Q crossing three difference media of refractive index `mu,2mu` and `3mu` respectively as shown in figure `phi_(P)` and `phi_(Q)` be the phase of the wave at points P and Q. Find the phase difference `phi_(Q)-phi_(P)`. [take `mu=1`]

A

0

B

`(pi)/(4)`

C

`(pi)/(2)`

D

`pi`

Text Solution

Verified by Experts

The correct Answer is:
C
Optical path difference between (OPD) P & Q
(O.P.D) `=2.25lamda_(0)xx1+(3.5lamda_(0))xx2+3lamda_(0)xx3=18.25lamda_(0)` and phase difference `Deltaphi=(2pi)/(lamda_(0))xxDeltax=(pi)/(2)`
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Knowledge Check

  • Light of wavelength lambda_(0) in air enters a medium of refractive index n. If two points A and B in this medium lie along the path of this light at a distance x, then phase difference phi_(0) between these two point is

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    `phi_(0) = (1)/(n) ((2 pi)/(lambda_(0))) x`
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  • Three charges +q, +q, +2q are arranged as shown in figure. What is the field at point P (center of side AC)

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  • Let P(a sec theta,b tan theta) and Q(a sec phi, b tan phi) where is the point of intersection of normals at P and Q then k is equal to

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