Q. 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

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
Explanation

(b) 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
$\mu_{1} = R.I.$ 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
$\mu = \frac{Real\ depth\ (h)}{Apparent\ depth\ (h')}$
$\Rightarrow 1.6 = \frac{0.04}{h'}$
$\Rightarrow h^{'} = 0.025\ m$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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