Following travelling electromagnetic wave Eₓ=0 Ey=E_0 sin (k x+omega t), Ez=-2 E_0 sin (k x-omega

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8.Electromagnetic waves

hard

The following travelling electromagnetic wave $E_x=0$ $E_y=E_0 \sin (k x+\omega t), E_z=-2 E_0 \sin (k x-\omega t)$ is

A

elliptically polarised

B

circularly polarised

C

linearly polarised

D

unpolarised

(KVPY-2010)

Solution

(c)

$E_y=E_0 \sin (k x+\omega t)$

and $\quad E_z=-2 E_0 \sin (k x-\omega t)$

As, sine function is variable for both components. So, in plane the tip of E-vector goes back and forth along a straight line.

So, correct option is $(c)$.

Standard 12

Physics

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