Question

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

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

Answer 1

The correct option is B A virtual image at 8cm

For refraction at spherical surface when the object is placed in denser medium, using the formula,

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

where ${\mu }_1$ is the refractive index of the medium into which the light rays are entering i.e. rarer medium

${\mu }_2$ is the refractive index of the medium from which the light rays are coming i.e. denser medium
$R$ is the radius of curvature of the spherical surface
$u$ and $v$ is the object distance and image distance respectively.

Given,

$u=-9cm$

${\mu }_1=1$ (air)
${\mu }_2=1.5$

$R=-12cm$

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

$\Rightarrow \frac{1}{v}=\frac{1-1.5}{-12}+\frac{1.5}{-9}$

$\Rightarrow \frac{1}{v}=\frac{-0.5}{-12}+\frac{-1.5}{9}=\frac{1}{24}-\frac{3}{18}=\frac{1}{24}-\frac{1}{6}$

$\Rightarrow \frac{1}{v}=\frac{1}{24}-\frac{1}{6}=\frac{1-4}{24}=\frac{-3}{24}$

$\Rightarrow v=\frac{-24}{3}=-8cm$

Thus, a virtual image, on the same side as object, is formed at a distance of $8cm$ from the pole.

Hence, the correct answer is OPTION B.