Consider a transparent hemisphere `(n=2)` in front of which a small object is placed in air `(n=1)` as shown in figure.
Q. What is the nature of final image of the object when `x=2R`?

Options

A. Erect and magnified
B. Inverted and magnified.
C. Erect and same size.
D. Inverted and same size.

Answer

d.

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^(@)`