Question

A convex lens refractive index $1.5$ has a focal length $18cm$ in air. Calculate it’s focal length when it is immersed in water of refractive index $\dfrac{4}{3}$. Take ${{f}_{1}}=18cm$.
$\begin{align}
  & \text{A}\text{. }{{f}_{1}} \\
 & \text{B}\text{. }2{{f}_{1}} \\
 & \text{C}\text{. 3}{{\text{f}}_{\text{1}}} \\
 & \text{D}\text{. }4{{f}_{1}} \\
\end{align}$

Answer 1

Hint: Use lens-makers formula to calculate the focal length in each case. Lens makers formula relates focal length with the refractive index of both lenses and the medium in which the lens is placed. As the radius of curvature remains constant when the lens is immersed in water so only the focal length of the lens is changed when it is immersed in water. After calculating the focal length in each case compare them to get the answer.

Formula used:
Lens makers formula is given by $\dfrac{1}{f}=\left[ \dfrac{{{\mu }_{2}}-{{\mu }_{1}}}{{{\mu }_{1}}} \right]\left[ \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right]$
Where $f=\text{ Focal length of lens}$,${{\mu }_{2}}=\text{ Refractive index of the lens}$,${{\mu }_{1}}=\text{ Refractive index of the medium in which lens is placed}$ ${{R}_{1}}\text{ and }{{R}_{2}}\text{ are the radius of curvature of both the curved surface of the lens}\text{.}$

Complete step by step answer:
Lens makers formula gives a relationship between focal length of lens and refractive index of lens material and the radius of curvature of the two surfaces of the lens.The lens-makers formula for a convex lens with focal length $f$ , radius of curvature of both the curved surface ${{R}_{1}}\text{ and }{{R}_{2}}$ and refractive index of lens ${{\mu }_{2}}$ when placed in the medium having refractive index ${{\mu }_{1}}$ is given by,

$\dfrac{1}{f}=\left[ \dfrac{{{\mu }_{2}}-{{\mu }_{1}}}{{{\mu }_{1}}} \right]\left[ \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right]$
Given the refractive index lens is, ${{\mu }_{2}}=1.5={{\mu }_{g}}$.

Case-I: When the lens is placed in air.

For air medium,${{\mu }_{1}}=1$
So focal length in this medium is given by $\dfrac{1}{{{f}_{air}}}=\left[ {{\mu }_{g}}-1 \right]\left[ \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right]$
Given ${{f}_{air}}={{f}_{1}}=18cm$.

Case-II: When the lens is immersed in water.

Refractive index of water, ${{\mu }_{w}}=\dfrac{4}{3}$.
In water medium the focal length of the lens is given by

$\dfrac{1}{{{f}_{water}}}=\left[ \dfrac{{{\mu }_{g}}-{{\mu }_{w}}}{{{\mu }_{w}}} \right]\left[ \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right]$
Now $\dfrac{{{f}_{water}}}{{{f}_{air}}}=\dfrac{\left[ {{\mu }_{g}}-1 \right]\left[ \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right]}{\left[ \dfrac{{{\mu }_{g}}-{{\mu }_{w}}}{{{\mu }_{w}}} \right]\left[ \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right]}$
$\dfrac{{{f}_{water}}}{{{f}_{air}}}=\dfrac{\left[ {{\mu }_{g}}-1 \right]}{\left[ \dfrac{{{\mu }_{g}}-{{\mu }_{w}}}{{{\mu }_{w}}} \right]}$

Putting the values of ${{\mu }_{g}}=1.5,{{\mu }_{w}}=\dfrac{4}{3}\text{ and }{{f}_{air}}={{f}_{1}}=18cm$ in above equation. Then

$\begin{align}
& \dfrac{{{f}_{water}}}{{{f}_{1}}}=\dfrac{\left[ 1.5-1 \right]}{\left[ \dfrac{1.5-\dfrac{4}{3}}{\dfrac{4}{3}} \right]}=\dfrac{0.5}{\left[ \dfrac{4.5-4}{4} \right]}=\dfrac{0.5}{\left[ \dfrac{0.5}{4} \right]}=0.5\times \dfrac{4}{0.5}=4 \\
 & \Rightarrow {{f}_{water}}=4{{f}_{1}} \\
\end{align}$

Hence, the correct answer is option D.

Note:
In ray optics, sign convention plays a major role. In every scenario the sign convention should be taken into account. In calculating the focal length and radius of curvature of a lens maker's formula the sign convention and refractive index of lens and medium should be taken into consideration.