Question

Two plano-concave lenses $(1 \,and\, 2)$ of glass of refractive index $1.5$ have radii of curvature $25\,cm$ and $20\,cm$. They are placed in contact with their curved surfaces towards each other and the space between them is filled with liquid of refractive index $ \frac{4}{3} $ . Then the combination is

• A. convex of focal length 70 cm
• B. concave of focal length 70 cm
• C. concave of focal length 66.6 cm
• D. convex of focal length 66.6 cm
• E. concave of focal length 72.5 cm

Answer 1

Answer: (C)

The focal length
$ \frac{1}{{{f}_{1}}}=(\mu -1)\frac{1}{-R} $ $ \frac{1}{{{f}_{1}}}=(1.5-1)\frac{1}{-25} $ $ {{f}_{1}}=-50\,cm $
The focal length $ \frac{1}{{{f}_{2}}}=(1.5-1)\times \frac{1}{-20} $
$ \frac{1}{{{f}_{2}}}=0.5\times \frac{1}{-20}m $
$ {{f}_{2}}=-\,40\,cm $ The focal length of bi-convex lens
$ \frac{1}{{{f}_{3}}}={{(}_{1}}{{n}_{2}}-1)\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right) $
$ \frac{1}{{{f}_{3}}}=\left( \frac{4}{3}-1 \right)\left( \frac{1}{20}+\frac{1}{25} \right) $
$ \frac{1}{{{f}_{3}}}=\frac{1}{3}\times \left( \frac{5+4}{100} \right) $
$ \frac{1}{{{f}_{3}}}=\frac{1}{3}\times \frac{9}{100} $
$ \frac{1}{{{f}_{3}}}=\frac{3}{100} $
$ \frac{1}{F}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}+\frac{1}{{{f}_{3}}} $
$ =-\frac{1}{50}-\frac{1}{40}+\frac{3}{100} $
$ \frac{1}{F}=\frac{-200}{3} $
$ F=-\,66.6\text{ }cm $
Negative sign represent the concave lens.