An equiconvex lens of glass $(_{a} μ_{g} = \frac{3}{2}),$ with each of its surfaces having the radius of curvature 20 cm, is placed with air in the object space and water in the image space.
Find the ratio of its focal length with reference to its face placed in air.

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Solution

Faced placed in air $= \frac{1}{f} = (\frac{μ_{2} - μ_{1}}{μ_{1}}) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) . .$

$R_{1} = + 20, R_{2} = - 20, μ_{2} = \frac{3}{2}, μ_{1} = 1.0, . . . .$
$f_{air} = 20 c m$
air in the object space and water in the image space
$\frac{1}{f} = (\frac{μ_{2} - μ_{1}}{μ_{1}}) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$R_{1} = + 20, R_{2} = - 20, μ_{2} = \frac{3}{2}, μ_{1} = μ_{w} = \frac{4}{3},$
$f_{w} = 40 c m$
ratio is $= \frac{40}{20} = 2$

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Similar questions

_{a} μ_{g} = \frac{3}{2} a n d_{a} μ_{w} = \frac{4}{3}

_{a} μ_{g} = \frac{3}{2} a n d_{a} μ_{w} = \frac{4}{3}

_{a} μ_{g} =

\frac{3}{2}

_{a} μ_{w} =

\frac{4}{3}

20 c m

\frac{5}{3},

40 cm

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