A convex lens of focal length 24 cm (μ = 1.5) is totally immersed in water (μ = 1.33). Find its focal length in water.

Solution
Focal length of lens in air (fₐ) = 24 cm
Refractive index of lens in air [latex](\mu_g)[/latex] = 1.5

We have, lens maker’s formula,

[latex]\frac{1}{f_a} = (\mu_g – 1)\left(\frac{1}{R_1} + \frac{1}{R_2}\right)[/latex]

[latex]\frac{1}{24} = (1.5 – 1)\left(\frac{1}{R_1} + \frac{1}{R_2}\right)[/latex]

[latex]\left(\frac{1}{R_1} + \frac{1}{R_2}\right) = \frac{1}{24\times 0.5} = \frac{1}{12}[/latex]

2nd Case (when immerged in water):
Focal length in water (f_w) = ?
Refractive index of water [latex](\mu_w)[/latex] = 1.33
Relative refractive index of glass w.r.t. water,

[latex]{}_w\mu_g = \frac{{}_a\mu_g}{{}_a\mu_w} = \frac{1.5}{1.33} = 1.127[/latex]

Again, using lens maker’s formula,

[latex]\frac{1}{f_w} = ({}_w\mu_g – 1)\left(\frac{1}{R_1} + \frac{1}{R_2}\right)[/latex]

Or, [latex]\frac{1}{f_w} = (1.127 – 1)\times\frac{1}{12}[/latex]
Or, [latex]f_w = \frac{12}{0.127} = 94.48[/latex] cm.

The image of an object is obtained on a screen with the help of a converging lens. The distance of the object is 40 cm from the screen and size of the image is 9 times the size of the object. Calculate distance of the screen from the lens and focal length of the lens.