Consider a transparent hemisphere `(n=2)` in front of which a small object is placed in air `(n=1)` as shown in figure.
Q. Consider a ray starting from O which strikes the spherical surface at grazing incident `(i=90^(@))`. Takin `x=R`, what will be the angle (from normal) at which the ray may emerge from the plane surface.

Options

A. `90^(@)`
B. `0^(@)`
C. `30^(@)`
D. `60^(@)`

Answer

a

Explanation

i. `(mu_(2))/(v)-(mu_(1))/(u)=(mu_(2)-mu_(1))/(R)`
Taking refraction first at curved surface,
`(2)/(v_(1))+(1)/(x)=(1)/(R)rArr v_(1)=(2Rx)/(x-R)`
For plane surface,
`v^(')=v_(1)RrArrv^(')=(xR+R^(2))/(x-R)`
`rArr (1)/(v)-(2(x-R))/(R(x+R))=0`
` (1)/(v)-(2(x-R))/(R(x+R))`
For virtual image,
`(1)/(v)lt0rArr (2(x-R))/(R(x+R))lt0`
`x lt R`
ii. For `x=2R`
`V_(1)=(4R^(2))/(R)= 4RrArru=-2R`
`m_(1)=(mu_(1))/(mu_(2)),(v)/(u)=(1)/(2), (4R)/((-2R))=-1`
`m_(2)=1rArr m_(1)m_(2)=-1`
Image is real, inverted, and of same size.
(iii) Hence, correct answer is `90^(@)`