A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index $n$ of…
A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index $n$ of the first lens is $1.5$ and that of the second lens is $1.2$. Both the curved surface are of the same radius of curvature $R=14$ $\mathrm{cm}$. For this bi-convex lens, for an object distance of 40 $\mathrm{cm}$, the image distance will be
$-280.0 \mathrm{~cm}$
$40.0 \mathrm{~cm}$
$21.5 \mathrm{~cm}$
$13.3 \mathrm{~cm}$
Solution
The focal length $\left(f_{1}\right)$ of the plano-cöneex Tens with $n=1.5$ using lens-maker formula
$\frac{1}{f_{1}}=\left(n_{1}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]=(1.5-1)\left[\frac{1}{14}-\frac{1}{\infty}\right]=\frac{1}{28}$
The focal length $\left(f_{2}\right)$ of the plano-convex lens with $n=1.2$ $\frac{1}{f_{2}}=\left(n_{2}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]=(1.2-1)\left[\frac{1}{\infty}-\frac{1}{-14}\right]=\frac{1}{70}$ Focal length F of the combination
$\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{20}$
Now, applying lens formula for the combination of lens
$\begin{array}{ll}
& \frac{1}{V}-\frac{1}{U}=\frac{1}{F} \Rightarrow \frac{1}{V}-\frac{1}{-40}=\frac{1}{20} \quad[\text { Given } \mu=40 \mathrm{~cm}] \\
\therefore \quad & V=40 \mathrm{~cm}
\end{array}$
