Optics Question 84
Question: A glass hemisphere of radius 0.04 m and R.I. of the material 1.6 is placed centrally over a cross mark on a paper (i) with the flat face; (ii) with the curved face in contact with the paper. In each case the cross mark is viewed directly from above. The position of the images will be
[ISM Dhanbad 1994]
Options:
A) (i) 0.04 m from the flat face; (ii) 0.025 m from the flat face
B) (i) At the same position of the cross mark; (ii) 0.025 m below the flat face
C) (i) 0.025 m from the flat face; (ii) 0.04 m from the flat face
D) For both (i) and (ii) 0.025 m from the highest point of the hemisphere
Show Answer
Answer:
Correct Answer: B
Solution:
Case (i) When flat face is in contact with paper.
$ \frac{{\mu _{2}}}{v}-\frac{{\mu _{1}}}{u}=\frac{{\mu _{2}}-{\mu _{1}}}{R} $
where $ {\mu _{2}} $ = R. I. of medium in which light rays are going = 1 $ {\mu _{1}} $ = R. I. of medium from which light rays are coming = 1.6 u = distance of object from curved surface = - 0.04 m R = - 0.04 m.
$ \therefore \frac{1}{v}-\frac{1.6}{(-0.04)}=\frac{1-1.6}{(-0.04)}\Rightarrow v=-0.04m $ i.e. the image will be formed at the same position of cross.
Case (ii) When curved face is in contact with paper $ \mu =\frac{Real\ depth\ (h)}{Apparent\ depth\ ({h}’)} $
$ \Rightarrow 1.6=\frac{0.04}{{{h}’}} $
$ \Rightarrow {h}’=0.025m $ (Below the flat face)
BETA
