Question

Focal length of a convex lens in air is 25 cm. If it is immersed in water, then calculate the
focal length of the lens. (n.w=4/3, n.g =3/2

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Answer 1

Answer:

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Answer 2

Answer:

The focal length of the lens when it is in water is 11.11cm.

Explanation:

Given:

$n_{g}=\frac{3}{2}$

$n_{w}=\frac{4}{3}$

$f_{a} = 25cm$

Here,

The refractive index of glass is denoted by $n_{g}$.

The refractive index of the air is denoted by $n_{a}$.

The refractive index of the water is denoted by $n_{w}$.

The focal length of a convex lens lense in the air is denoted by  $f_{a}$.

The focal length of a convex lens in the water is denoted by  $f_{w}$.

The radius of curvature of one side of a lens is denoted by $R_{1}$.

The radius of curvature of the second side of a lens is denoted by $R_{2}$.

Now,

When the lens is in the air,

By the equation,

$\frac{1}{f_{a} }=( \frac{n_{g} }{n_{a} } -1)(\frac{1}{R_{1} }- \frac{1}{R_{2} } )\\\frac{1}{25}=( \frac{\frac{3}{2} }{1} -1)(\frac{1}{R_{1} }- \frac{1}{R_{2} } )$

$(\frac{1}{R_{1} }- \frac{1}{R_{2} } )=\frac{2}{25}$

Then,

When the lens is in the water,

By the equation,

$\frac{1}{f_{w} }=( \frac{n_{g} }{n_{w} } -1)(\frac{1}{R_{1} }- \frac{1}{R_{2} } )\\\frac{1}{f_{w} }=( \frac{\frac{3}{2} }{\frac{4}{3} } -1)(\frac{1}{R_{1} }- \frac{1}{R_{2} } )\\\frac{1}{f_{w} }=( \frac{\frac{3}{2} }{\frac{4}{3} } -1)\frac{2}{25}$ ∴ $(\frac{1}{R_{1} }- \frac{1}{R_{2} } )=\frac{2}{25}$

$f_{w} =11.11cm$

So, the focal length of the lens when it is in water is 11.11cm.