A mark placed on the surface of a sphere is viewed through glass from a position directly opposite. If the diameter of the sphere is 10 cm and refractive index of glass is 1.5, the image would be at
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → -20 cm
The mark is on the far surface of the glass sphere and the observer looks through the near surface, so the light travels from the glass into air and is refracted at a single spherical surface. The formula for refraction at a spherical surface, light going from a medium of index \( n_1 \) to one of index \( n_2 \), is
\( \displaystyle \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \)
Sign convention: distances are measured from the near surface, positive along the direction of the light (from the mark towards the observer).
\( n_1 = 1.5 \) (glass), \( n_2 = 1 \) (air)
\( u = -10\,\mathrm{cm} \) (the mark is one diameter behind the near surface)
\( R = -5\,\mathrm{cm} \) (the centre of the sphere lies on the side of the object)
\( \displaystyle \frac{1}{v} - \frac{1.5}{-10} = \frac{1 - 1.5}{-5} \)
\( \displaystyle \frac{1}{v} + 0.15 = 0.1 \Rightarrow \frac{1}{v} = -0.05 \)
\( v = -20\,\mathrm{cm} \)
The negative sign means a virtual image, 20 cm from the near surface on the same side as the mark, inside the glass.