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Q.

The radii of curvatures of convex surface, concave surface of a concavo-convex lens are in the ratio 1 : 2. The refractive index of the material of lens is 5/4 and its focal length is 40 cm. This lens is cut into a plano-convex lens and plano-concave lens. If the plano convex lens is immersed in water then its focal length is

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a

+ 20 cm

b

-20 cm

c

+ 80 cm

d

-80 cm

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{1}{f} = (μ - 1) (\frac{1}{R} - \frac{1}{2 R}) \\ \frac{1}{40} = (\frac{5}{4} - 1) (\frac{1}{2 R}) \\ R = 5 c m \\ \frac{1}{f_{2}} = (\frac{μ}{μ_{w}} - 1) (\frac{1}{R}) \\ \frac{1}{f_{2}} = (\frac{5 / 4}{4 / 3} - 1) (\frac{1}{5}) = - \frac{1}{80} \\ f_{2} = - 80 c m \end{array}$

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The radii of curvatures of convex surface, concave surface of a concavo-convex lens are in the ratio 1 : 2. The refractive index of the material of lens is 5/4 and its focal length is 40 cm. This lens is cut into a plano-convex lens and plano-concave lens. If the plano convex lens is immersed in water then its focal length is