Question

A convex refracting surface of radius of curvature `20 cm` separates two media of refractive indices `4//3 and 1.60`. An object is placed in the first medium `(mu = 4//3)` at a distance of `200 cm` from the refracting surface. Calculate the position of image formed.

Answer 1

Correct Answer - At `234.15 cm` in denser medium
Here, `R = 20 cm, mu_1= 4 //3, mu_2 = 1.60`
`u = -200 cm, v = ?`
As refraction occurs from rarer to denser medium,
therefore, `-(mu_1)/(u)+(mu_2)/(v)=(mu_2 - mu_1)/( R)`
-`(4//3)/(-200)+(1.60)/(v) = (1.6 - 4//3)/(20)`
`(1)/(150)+(8)/(5 v)=(0.27)/(20)`
`(8)/(5 v) = (0.27)/(20)-(1)/(150) = (4.05 - 2)/(300)`
`5 v xx 2.05 = 2400`
`v = (2400)/(5 xx 2.05) = 234.15 cm`
(in denser medium).