A convex lens (n=2) is submerged in water that was previously suspended in the air. What would be the change in the focal length?

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Solution

Given:
Refractive index of the surrounding, = $n_{1}$
Refractive index the lens, = $n_{2}$
Using lens maker's formula,
$\frac{1}{f} = (\frac{n_{2}}{n_{1}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}$ )

focal length when lens is placed in air:
$n_{1} = 1$
$n_{2} = 2$
$\frac{1}{f_{1}} = (\frac{n_{2}}{n_{1}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}$ )
Substituting the values
$\frac{1}{f_{1}} = (\frac{2}{1} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}$ )
$\frac{1}{f_{1}} = (1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}$ ) ......(1)

focal length when lens is placed in water:
$n_{1} = 1.33$
$n_{2} = 2$
$\frac{1}{f_{2}} = (\frac{n_{2}}{n_{1}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}$ )
Substituting the values
$\frac{1}{f_{2}} = (\frac{2}{1.33} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}$ )
$\frac{1}{f_{2}} = (0.503) (\frac{1}{R_{1}} - \frac{1}{R_{2}}$ ) .......(2)

Dividing eq(1) and eq(2)
$\frac{1}{f_{1}} \times \frac{f_{2}}{1} = \frac{1}{0.503}$
$\frac{f_{2}}{1} = \frac{f_{1}}{0.503}$
$f_{2} \approx 2 f_{1}$
There will be an increase in focal length of the convex lens. This is because the refractive index of glass with respect to water is less than the refractive index of glass with respect to air.

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