Urgent! Q4. According to the solution I've understood how r and 2r came but I don't understand that 2r = 60 degrees??? At first light would rotate clockwise by an angle 60-r and at second surface it would again rotate clockwise by r so net rotation will be 60 hence parallel to AB. But why r has to be 30 degrees here? And this question is correct.

4. A ray of light falls on a transparent sphere with centre at C as shown in figure. The ray emerges from the sphere 
parallel to line AB. The refractive index the sphere is:- 


(A) $\sqrt{2}$
(B) $\sqrt{3}$
(C) 3/2 
(D) 1/2
 

Dear Student,


imagine the following picture,

point 1(P)  refraction at 1st surface
point 2 (Q) refraction at 2nd surface
point 3 (C) if you extend the normal at surface 2 till it meets the horizontal line ( it will be the centre)

So in triangle PQC  angle QPC= angle CQP  ( side PC and Side QC are equal (= radius) [ isosceles triangle ] 
These angles are also the angle of refraction and internal angle of incidence respectiveley at surface 1 and 2

Thus obviously angle of emergence has to equal to initial angle of incidence


Now using property of parallel lines, external angle to triangle PQC is also equal to 60 

but ​angle QPC+ angle CQP= external angle

or 2 x angle of refraction = 60  or angle of refraction (r) = 30 

refractive index= $\frac{\sin i}{\sin r}\Rightarrow \frac{\sin 60}{\sin 30}\Rightarrow \frac{{\displaystyle \frac{\sqrt{3}}{2}}}{{\displaystyle \frac{1}{2}}}\Rightarrow \sqrt{3} \left(ans\right)$

Regards.