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.

Answer 1

Step 1: Given that:

The refractive index of denser medium= $1.5$

Radius of curvature($R$) = $12cm$

Object distance($u$)= $9cm$

Step 2: Calculation of image distance:

According to the refraction formula from denser to medium to rare medium,

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

Where,

${\mu }_1$ = Refractive index of rare medium

${\mu }_2$ = Refractive index of denser medium

$v=Imagedistance$

Thus, putting all the values, we get

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

$\frac{1}{v}+\frac{1}{6}=\frac{-0.5}{-12}$

$\frac{1}{v}=\frac{1}{24}-\frac{1}{6}$

$\frac{1}{v}=\frac{1-4}{24}$

$\frac{1}{v}=\frac{-3}{24}$

$v=-8cm$

Thus, the image distance will be $8cm$ .