Group I Group II P. Ex = E0/√2cos(ωt + kz) Ey = E0sin(ωt + kz + π) 1. Circularly Polarized

Question

Group I contains x - and y - components of the electric field and Group II contains the type of polarization of light.

Group I	Group II
P.   Ex = E0/√2cos(ωt + kz)        Ey = E0sin(ωt + kz + π)	1.  Circularly Polarized
Q.  Ex = E0sin(ωt + kz + π/6)      Ey = E0sin(ωt + kz + π/3)	2.  Elliptically Polarized
R.  Ex = E0sin(ωt + kz + 3π/4)         Ey = E0sin(ωt + kz + π/4)	3. Linearly Polarized

The correct set of matches is

 (a)  P →1;Q → 2;R → 3

(b)  P →1;Q → 3;R → 2

(c)  P → 2;Q →1;R → 3

(d)  P → 2;Q → 3;R →1

Answer 1

Correct option (c) P → 2;Q →1;R → 3 

Explanation:

The phase difference between E_x and E_y is  π/2 with different amplitude. Therefore the resultant will be elliptically polarized. 

The phase difference between E_x and E_y is π/2 with same amplitude. Therefore the resultant will be circularly polarized.

The phase difference between E_x and E_y is π with different amplitude. Therefore the resultant will be linarly polarized.