AP EAMCET201921 Apr 2019Morning ShiftPhysicsWave OpticsActual
Unpolarised light from air incidents on the surface of a transparent medium of refractive index 1.414 such that the reflected light is completely polarised. Match the angles given in List-I with the corresponding values given in List-II. The correct match is Codes array llll A & B & C & D array
Options
- A(ii) (iii) (i) (iv)
- B(ii) (iii) (iv) (i)
- C(iv) (i) (iii) (ii)
- D(iv) (iii) (i) (ii)
Correct answer
D. (iv) (iii) (i) (ii)
Step-by-step solution
Figure showing the ray diagram of refraction at a plane surface. Given, i_p= angle of polarisation So, i_p= ⁻¹ i_p= ⁻¹ 2 = ⁻¹ 2 3 From Snell's law, i_p= r Here, =1.414= 2 r= i_p aligned r & = 1 2 2 3 r & = ⁻¹ 1 3 aligned angle of refraction Now, the angle of deviation, aligned & =i_p-r & = ⁻¹ 2 3 - ⁻¹ 1 3 aligned Angle of reflection, gathered =90^ -r r=90^ - ⁻¹ 1 3 =90^ - 1 3 = (90^ - ) = 1 3 = ⁻¹ 1 3 gathered Angle between incident and completely polarised light is given by aligned & =i_p+ & = ⁻¹ 2 3 + ⁻¹ ( 1 3 )=