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Question: A thin symmetrical double convex lens of refractive index \[{\mu _2} = \]1.5 is placed between a med...

A thin symmetrical double convex lens of refractive index μ2={\mu _2} = 1.5 is placed between a medium of refractive index μ1={\mu _1} = 1.4 to the left and another medium of refractive index μ3={\mu _3} = 1.6 to the right. Then, the system behaves as.
A. A convex lens
B. A concave lens
C. A glass plate
D. A convo concave lens

Explanation

Solution

We can consider the refraction through the lens as a refraction through a spherical surface twice i.e. once from the front side of the lens and then from the second circular surface of the lens. Refraction through a circular surface is given as−μ1u+μ1v=μ2−μ1R - \dfrac{{{\mu _1}}}{u} + \dfrac{{{\mu _1}}}{v} = \dfrac{{{\mu _2} - {\mu _1}}}{R}.
If the rays coming from infinity meet at infinity the lens acts as a glass plate.

Complete step by step answer:
Let us consider two cases for refraction through the two surfaces of the lens. We know that the equation for refraction through a spherical surface is given by as-−μ1u+μ1v=μ2−μ1R - \dfrac{{{\mu _1}}}{u} + \dfrac{{{\mu _1}}}{v} = \dfrac{{{\mu _2} - {\mu _1}}}{R} -----------(1)
Case 1: When the rays travel from first medium(μ1=1.4)\left( {{\mu _1} = 1.4} \right) to the first surface of the lens(μ2=1.5)\left( {{\mu _2} = 1.5} \right).
Substituting the values in equation 1,
\-1.4−∞+1.5v=1.5−1.4R ⇒0+1.5v=0.1R ⇒v=15R  \- \dfrac{{1.4}}{{ - \infty }} + \dfrac{{1.5}}{v} = \dfrac{{1.5 - 1.4}}{R} \\\ \Rightarrow 0 + \dfrac{{1.5}}{v} = \dfrac{{0.1}}{R} \\\ \Rightarrow v = 15R \\\
Case 2: When the rays travel from the second surface of the lens(μ1=1.5)\left( {{\mu _1} = 1.5} \right) to the second medium(μ2=1.6)\left( {{\mu _2} = 1.6} \right). The image distance found in case 1 is used as object distance for this case, u=15R = 15R.
Substituting the values in equation 1.
\-1.515R+1.6V=1.6−1.5−R ⇒−0.1R+1.6V=−0.1R ⇒1.6V=0 ⇒1V=0⇒V=∞  \- \dfrac{{1.5}}{{15R}} + \dfrac{{1.6}}{V} = \dfrac{{1.6 - 1.5}}{{ - R}} \\\ \Rightarrow - \dfrac{{0.1}}{R} + \dfrac{{1.6}}{V} = - \dfrac{{0.1}}{R} \\\ \Rightarrow \dfrac{{1.6}}{V} = 0 \\\ \Rightarrow \dfrac{1}{V} = 0 \Rightarrow V = \infty \\\
Therefore the final image distance is V=∞V = \infty .

Hence the correct option is C the system acts as a glass plate.

Note: Refraction through a lens with two different mediums on its either side can be considered as refraction from a spherical surface twice. The sign of object distance, image distance and focal length should be taken carefully according to the convention to avoid error in answer.