Question

A double convex lens of focal length $6cm$ is made of glass of refractive index $1.5$. The radius of curvature of one surface is double that of another surface. The value of small radius of curvature is:
$\begin{align}
  & \text{A}\text{. }6cm \\
 & \text{B}\text{. }4.5cm \\
 & \text{C}\text{. }9cm \\
 & \text{D}\text{. }4cm \\
\end{align}$

Answer 1

Hint: There can be different types of optical lenses, based on the classification of their refracting surfaces. Lens maker formula is used to calculate the focal length of different lenses. For calculating the radius of curvature of the smaller lens, we will apply lens maker formula on the given biconvex lens.

Complete step by step answer:
A lens is a transmissive optical instrument that can focus or disperse an incident light beam using the phenomenon of refraction of light. A simple lens consists of a single piece of transparent material, while a compound lens consists of several simple lenses or elements usually arranged along a common axis.
Lenses are classified on the basis of the curvature of the two optical surfaces. A lens is Biconvex, or double convex, if both the surfaces are convex. If both the surfaces have the same radius of curvature, the lens is called an Equiconvex lens. A lens having two concave surfaces is Biconcave, or double concave lens. If one of the surfaces of the lens is flat, the lens is called as Plano-convex or Plano-concave, depending upon the curvature of the other surface. A lens having one convex and one concave side is called a convex-concave lens.
Let us consider a thin lens with two refracting surfaces having radii of curvatures ${{R}_{1}}$ and ${{R}_{2}}$. Let the refractive index of the surrounding be ${{n}_{1}}$ and that of the lens be ${{n}_{2}}$.

With the help of formula for refraction at a single spherical surface,
For the first surface, we have,
$\dfrac{{{n}_{2}}}{{{v}_{1}}}-\dfrac{{{n}_{1}}}{u}=\dfrac{{{n}_{2}}-{{n}_{1}}}{{{R}_{1}}}$
For the second surface, we have,
$\dfrac{{{n}_{1}}}{v}-\dfrac{{{n}_{2}}}{{{v}_{1}}}=\dfrac{{{n}_{1}}-{{n}_{2}}}{{{R}_{2}}}$
Adding the above two equations,
$\dfrac{{{n}_{1}}}{v}-\dfrac{{{n}_{1}}}{u}=\left( {{n}_{2}}-{{n}_{1}} \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$
$\dfrac{1}{v}-\dfrac{1}{u}=\left( \dfrac{{{n}_{2}}}{{{n}_{1}}}-1 \right)\left[ \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right]$
When $u=\infty $ and $v=f$
We get,
\[\dfrac{1}{f}=\left( \dfrac{{{n}_{2}}}{{{n}_{1}}}-1 \right)\left[ \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right]\]
Therefore,
\[\dfrac{1}{f}=\left( \mu -1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)\]
Where, $\mu $ is the refractive index of the material.
The above equation is known as Lens maker formula.
We are given that a double convex lens has focal length $6cm$ and refractive index $1.5$.
Also,
$\begin{align}
  & {{R}_{1}}=R \\
 & {{R}_{2}}=-2R \\
\end{align}$
Using lens maker formula,
$\dfrac{1}{f}=\left( \mu -1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$
Putting values,
$\begin{align}
  & f=6cm \\
 & \mu =1.5 \\
\end{align}$
We get,
$\begin{align}
  & \dfrac{1}{6}=\left( 1.5-1 \right)\left[ \dfrac{1}{R}-\dfrac{1}{\left( -2R \right)} \right] \\
 & \dfrac{1}{6}=\dfrac{0.5\times 3}{2R} \\
 & R=4.5cm \\
\end{align}$
The radius of curvature of the smaller lens is $4.5cm$

So, the correct answer is “Option B”.

Note: While deriving the lens formula, it should be remembered that the image distance from the refraction at first surface becomes the object distance for the refraction at second surface since the light ray is passing undisturbed from the lens. Also, the refractive indices of the two spherical materials, of which lens is a part of, can also be different. In the above question, the refractive indices of both the materials were the same as the refractive index of glass.