Question

(i) A point object is placed on the principal axis of a convex spherical surface of radius of curvature R, which separated the two media of refractive indices n₁ and n₂ (n₂ > n₁). Draw the ray diagram and deduce the relation between the object distance (u), image distance (v) and the radius of curvature (R) for refraction to take place at the convex spherical surface from rarer to denser medium.

(ii) A converging lens has a focal length of 20 cm in air. It is made of material of refractive index 1.6. If it is immersed in a liquid of refractive index 1.3, find its new focal length.

Answer 1

(i)

For small angles

tan ∠NOM = \(\frac{MN}{OM}\) : tan ∠NCM = \(\frac{MN}{NC}\) and tan ∠NIM = \(\frac{MN}{MI}\)

For ∆ NOC, i is exterior angle, therefore

i = ∠NOM + ∠NCM = \(\frac{MN}{CM}+\frac{MN}{NC}\)

Similarly, r = \(\frac{MN}{MC}-\frac{MN}{MI}\)

For small angles Snells law can be written as n₁i = n₂r

\(∴\frac{n_1}{OM}-\frac{n_2}{MI}=\frac{n_2-n_1}{R}\)

∴ OM = -u, MI = +v MC = +R (using sign conversion)

∴ \(\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}\)

(ii) Lens marker formula is

\(\frac{1}{f_0}=(\frac{n_2}{n_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})\)

∴ \(\frac{1}{20}=(1.6-1)(\frac{1}{R_1}-\frac{1}{R_2})\)

∴ \((\frac{1}{R_1}-\frac{1}{R_2})=\frac{1}{20\times0.6}=\frac{1}{12}\)

Let f be the focal length of the lens in water

∴ \(\frac{1}{f'}=(\frac{1.6-1.3}{1.3})(\frac{1}{R_1}-\frac{1}{R_2})=\frac{0.3}{12\times1.3}\)

or \(f'=\frac{120\times1.3}{3}=52\,cm\)