A convex lens made of glass (refractive index = 1.5 ) has focal length 24 cm in air. When it is totally immersed in wate…

The focal length in air is given by the lens formula:

\(\frac{1}{f}=\left({\mu }_{l}-1\right)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)\)

where \(\mu\), is the refractive index of the lens material ( 1.5 for glass), and \({R}_{1}\) and \({R}_{2}\) are the radii of curvature of the lens surfaces. For air, the focal length f = 24 cm, So, \(\frac{1}{24}=(1.5-1)\cdot \left(\frac{2}{R}\right)\)

When immersed in water.

\(\frac{1}{{f}^{'}}=\left(\frac{{\mu }_{l}}{{\mu }_{s}}-1\right)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)\)

Substituting the values for \({\mu }_{l}=1.5\), \({\mu }_{s}=1.33\) and the same \({R}_{1},{R}_{2}\) from before:

\(\frac{1}{{f}^{'}}=\left(\frac{1.5}{1.33}-1\right)\left(\frac{2}{R}\right)\)

Dividing the two equations for the focal lengths in air and in water, we get:

\(\frac{{f}^{'}}{24}=\frac{\left(1.5-1\right)}{\left(\frac{1.5}{1.33}-1\right)}\\ f'=96cm\)

Thus, the new focal length in water is:

\({f}^{'}=96cm\)