Two Plano-concave lens of glass of refractive index 1.5 have radii of curvature 20 cm and 30 cm. They are placed in contact with curved surfaces towards each other and the space between them is filled with a liquid of refractive index 4/3. Find the focal length of the system.

Solution

As shown in the figure, the system is equivalent to combination of three lenses in contact,

Refractive index,

μ

= 1.5
Refractive index of liquid =

\frac{4}{3}

Radius of curvature of first lens, R₁= 20 cm
Radius of curvature of second lens, R₂ = 30 cm

We know,

\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} + \frac{1}{f_{3}}

By lens makers formula

\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})

Now,

\frac{1}{f_{1}} = (\frac{3}{2} - 1) [\frac{1}{\infty} - \frac{1}{20}] = \frac{1}{40} cm

\frac{1}{f_{2}} = (\frac{4}{3} - 1) [\frac{1}{20} - \frac{1}{- 30}] = \frac{5}{180} cm

\frac{1}{f_{3}} = (\frac{3}{2} - 1) [\frac{1}{- 30} - \frac{1}{\infty}] = - \frac{1}{60} cm

∴

\frac{1}{F} = - \frac{1}{40} + \frac{5}{180} - \frac{1}{60}

\Rightarrow

F = - 72 cm .

Thus, the system will behave as a concave lens of focal length 72 cm.

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