Question

Two plano-concave lenses $(1and2)$ of glass of refractive index $1.5$ have radii of curvature $25cm$ and $20cm$. 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
$\frac{1}{f_1}=(\mu -1)\frac{1}{-R}$ $\frac{1}{f_1}=(1.5-1)\frac{1}{-25}$ $f_1=-50cm$
The focal length $\frac{1}{f_2}=(1.5-1)\times \frac{1}{-20}$
$\frac{1}{f_2}=0.5\times \frac{1}{-20}m$
$f_2=-40cm$ The focal length of bi-convex lens
$\frac{1}{f_3}={(}_1{n}_2-1)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)$
$\frac{1}{f_3}=\left(\frac{4}{3}-1\right)\left(\frac{1}{20}+\frac{1}{25}\right)$
$\frac{1}{f_3}=\frac{1}{3}\times \left(\frac{5+4}{100}\right)$
$\frac{1}{f_3}=\frac{1}{3}\times \frac{9}{100}$
$\frac{1}{f_3}=\frac{3}{100}$
$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}$
$=-\frac{1}{50}-\frac{1}{40}+\frac{3}{100}$
$\frac{1}{F}=\frac{-200}{3}$
$F=-66.6cm$
Negative sign represent the concave lens.