Question

An equi-convex lens $(\mu =1.5)$ is combined with an equi-concave lens $(\mu =1.3)$. The radii of curvature of all surfaces is the same. (Assume thin lenses)

• A. The combination behaves like a converging lens in a medium of refractive index 1.4 and diverging lens in a medium of refractive index 1.6.
• B. The combination behaves like a converging lens in a medium of refractive index 1.2 as well as 1.4.
• C. The combination behaves like a diverging lens in a medium of refractive index 1.2 as well as 1.6
• D. The combination behaves like a diverging lens in a medium of refractive index 1.4 and converging lens in a medium of refractive index 1.6.

Answer 1

The correct option is B The combination behaves like a converging lens in a medium of refractive index 1.2 as well as 1.4.



For medium ${\mu }_1$ and ${\mu }_m$
by using lens formula
$\frac{{\mu }_1}{v_1}-\frac{{\mu }_m}{u}=\frac{{\mu }_1-{\mu }_m}{R}$ ....(1)

For medium ${\mu }_2$ and ${\mu }_1$
$\frac{{\mu }_2}{v_2}-\frac{{\mu }_1}{v_1}=\frac{{\mu }_2-{\mu }_1}{-R}$ ....(2)

For medium ${\mu }_m$ and ${\mu }_2$
$\frac{{\mu }_m}{v}-\frac{{\mu }_2}{v_2}=\frac{{\mu }_m-{\mu }_2}{+R}$ ....(3)

On solving eq. (1), eq.(2), and eq.(3)
$\frac{{\mu }_m}{v}-\frac{{\mu }_m}{u}=({\mu }_1-{\mu }_2)\frac{2}{R}$
$\frac{1}{v}-\frac{1}{u}=\left(\frac{{\mu }_1-{\mu }_2}{{\mu }_m}\right)\frac{2}{R}$

using lens formula, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
So, $\frac{1}{f_{equ}}=\left(\frac{{\mu }_1-{\mu }_2}{{\mu }_m}\right)\frac{2}{R}\frac{1}{f_{equ}}=(\frac{0.2}{{\mu }_m}.\frac{2}{R})$
$\frac{1}{f_{eq}}=\frac{0.4}{R.{\mu }_m}$
As the final expression $f_{eq}$ is $+ve$ in nature.
Hence, the combination will always behave like a converging lens irrespective of the refractive index of the medium.