question_answer

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]

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

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. \[\therefore \frac{1}{v}-\frac{1.6}{(-0.04)}=\frac{1-1.6}{(-0.04)}\Rightarrow v=-\,0.04\,m\] i.e. the image will be formed at the same position of cross. Case (ii) When curved face is in contact with paper \[\mu =\frac{Real\ depth\ (h)}{Apparent\ depth\ ({h}')}\]          \[\Rightarrow 1.6=\frac{0.04}{{{h}'}}\]\[\Rightarrow {h}'=0.025\,m\]       (Below the flat face)