Question

A denser medium of refractive index 1.5 has a concave surface of radius of curvature 12 cm. An object is situated in the denser medium at a distance of 9 cm from the pole. Locate the image due to refraction in air

• A. A real image at 8 cm
• B. A virtual image at 8 cm
• C. A real image at 4.8 cm
• D. A virtual image at 4.8 cm

Answer 1

The correct option is B A virtual image at 8 cm

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_{2-}{\mu }_1}{R}$

$\frac{1}{V}-\frac{1.5}{-9}=\frac{1-1.5}{-12}$

$\frac{1}{V}=\frac{1}{24}-\frac{3}{18}$

$\Rightarrow V=-8cm$