The focal length of a glass convex lens in air is f a and that in water is f w . The refractive index from air to glass is 2 3 and that from air to water is 3 4 . The ratio f a : f w is

Question

The focal length of a glass convex lens in air is f a ​ and that in water is f w ​ . The refractive index from air to glass is 2 3 ​ and that from air to water is 3 4 ​ . The ratio f a ​ : f w ​ is

TestHub · Accepted Answer

Using the Lens Maker's formula for the lens in air:

$\frac{1}{f _{a}} = (μ_{g} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

Substituting $μ_{g} = \frac{3}{2}$ :

$\frac{1}{f _{a}} = (\frac{3}{2} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) = \frac{1}{2} (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

For the lens in water, the relative refractive index is $\frac{μ _{g}}{μ _{w}}$ :

$\frac{1}{f _{w}} = (\frac{μ _{g}}{μ _{w}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

Substituting $μ_{g} = \frac{3}{2}$ and $μ_{w} = \frac{4}{3}$ :

$\frac{1}{f _{w}} = (\frac{3/2}{4/3} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) = (\frac{9}{8} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) = \frac{1}{8} (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

Dividing the equation for $f_{w}$ by the equation for $f_{a}$ :

$\frac{f _{w}}{f _{a}} = \frac{1/2}{1/8} = 4$

$\Rightarrow \frac{f _{a}}{f _{w}} = \frac{1}{4}$

The ratio $f_{a} : f_{w}$ is $1 : 4$ .

Answer: $1 : 4$

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