Focal length of a convex lens in air is 10 cm. Find its focal length in water. Given that µg = 3/2 and µw = 4/3.

Steps to Solve

1. Understanding the Lens Maker's Formula The lens maker's formula relates the focal length of a lens in different media. The formula is given as: $$ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) $$ Where ( f ) is the focal length, ( \mu ) is the refractive index, and ( R_1 ) and ( R_2 ) are the radii of curvature of the lens surfaces.

2. Applying the Formula in Air and Water The focal length in air ( f_{air} ) is 10 cm, where the refractive index ( \mu_{glass} = \frac{3}{2} ): $$ \frac{1}{f_{air}} = \left( \frac{3}{2} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) $$ This gives: $$ \frac{1}{10} = \frac{1}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) $$

3. Finding the New Focal Length in Water In water, the refractive index is ( \mu_{water} = \frac{4}{3} ): $$ \frac{1}{f_{water}} = \left( \frac{3}{2} - \frac{4}{3} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) $$ We calculate:

• First find ( \frac{3}{2} - \frac{4}{3} ): $$ \frac{3}{2} = \frac{9}{6}, \quad \frac{4}{3} = \frac{8}{6} $$ Thus, $$ \frac{3}{2} - \frac{4}{3} = \frac{9}{6} - \frac{8}{6} = \frac{1}{6} $$

4. Relating Focal Lengths Since ( \frac{1}{f_{air}} = \frac{1}{10} ), for ( f_{water} ): $$ \frac{1}{f_{water}} = \frac{1}{6} \left( \frac{1}{2} \cdot \frac{1}{10} \right) = \frac{1}{60} $$ Thus, $$ f_{water} = 60 \text{ cm} $$