Question

Two plano-concave lenses of glass of refractive index 1.5 have radii of curvatures 20 cm, 30 cm.They are placed in contact with curved surface towards each other and the space between them is filled with a liquid of refractive index 4/3. The focal length of the system is :

• A. $48cm$
• B. $72cm$
• C. $12cm$
• D. $24cm$

Answer 1

Answer: (B)

$72cm$

Given,

Refractive index of the lens, ${\mu }_g=1.5$

Radius of curvature of first plano-concave lens, $R_1=20cm$

Radius of curvature of second plano-concave lens, $R_2=30cm$

Refractive index of the liquid ${\mu }_l=\frac{4}{3}$

Using Lens makers formula,

$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

Now, the system can be considered as 3 lenses in contact as shown in the figure.

Focal length of lens 1

$\frac{1}{f_1}=(\frac{3}{2}-1)(\frac{1}{\infty }-\frac{1}{20})=(\frac{1}{2})(\frac{-1}{20})=\frac{-1}{40}$

$f_1=-40cm$

Focal length of lens 2

$\frac{1}{f_2}=(\frac{4}{3}-1)(\frac{1}{20}-\frac{1}{-30})=(\frac{1}{3})(\frac{5}{60})=\frac{5}{180}$

$f_2=36cm$

Focal length of lens 3

$\frac{1}{f_3}=(\frac{3}{2}-1)(\frac{1}{-30}-\frac{1}{\infty })=(\frac{1}{2})(\frac{-1}{30})=\frac{-1}{60}$

$f_3=-60cm$

The equivalent focal length of the combination is given as

$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}$

$\frac{1}{F}=\frac{-1}{40}+\frac{1}{36}+\frac{-1}{60}$

$\frac{1}{F}=\frac{-9+10-6}{360}=\frac{-5}{360}$

$F=\frac{360}{-5}=-72cm$

$\therefore$ The system will behave as a Concave Lens of focal length $72cm$.

Hence, the correct answer is OPTION B.