Question

A plane convex lens is made of material having refractive index $1.5$. The radius of curvature of the curved surface is $60\;{\text{cm}}$. The focal length of the lens is _____ ${\text{cm}}$.
A. $- 60$
B. $120$
C. $60$
D. $- 120$

Answer 1

Note that the radius of curvature of the plane surface is infinity. To obtain the relation between focal length, refractive index and radius of curvatures, use the lens formula

Complete step by step answer:
Given,
Refractive index, $\mu = 1.5$
Radius of curvature of curved surface, ${R_2} = - 60\;{\text{cm}}$
Radius of curvature of the plane surface is infinite.
i.e. ${R_1} = \propto$

Now, from the lens formula, the relation between focal length, refractive index and radius of curvatures is given by,
$\dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ …… (1)

Substitute the values of $\mu = 1.5$,${R_2} = - 60\;{\text{cm}}$, and ${R_1} = \propto$in equation (1).
\dfrac{1}{f} = \left( {1.5 - 1} \right)\left( {\dfrac{1}{ \propto } - \dfrac{1}{{\left( { - 60} \right)}}} \right) \\\ \dfrac{1}{f} = 0.5 \times \dfrac{1}{{60}} \\\ \dfrac{1}{f} = \dfrac{1}{{120}} \\\ f = 120 \\\

Therefore, the required focal length is $120\;{\text{cm}}$.

So, the correct answer is “Option B”.

Additional Information:
The centre of curvature of a spherical lens and mirror surface is located either around or decentralised from the local optical axis of the system. Mostly on the local optical axis, the vertex of its lens surface is located. The radius of curvature of a surface seems to be the distance from vertex to the centre of curvature.

In optics, a material's refractive index is a dimensionless number that defines how easily light passes through the object. It is identified as $n = cv$, where the velocity of light throughout the vacuum is $c$ and the phase velocity of light throughout the medium is $v$.
The plane mirror can be called an aspect of an unimaginably huge spherical surface. The bending radius is infinity on an infinitely large spherical surface.

Note:
To calculate the focal length we use the lens formula that gives the relation between focal length, radius of curvature and refractive index as well. We consider the radius of curvature of the plane surface as ${R_1}$ and that of the curved surface is ${R_2}$. We subtract $\dfrac{1}{{{R_2}}}$ from $\dfrac{1}{{{R_1}}}$ instead of $\dfrac{1}{{{R_1}}}$from $\dfrac{1}{{{R_2}}}$.