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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

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

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

Regards.

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