A convex lens of focal length 24\ cm and of refractive index 1.5 is totally immersed in water of refractive index 1.33 . Find its focal length in water.

To determine the new focal length of the convex lens when immersed in water, we use the lensmaker’s equation and the relationship between refractive indices.

Step 1: Recall the lensmaker’s equation in air

The focal length $f$ of a lens in air (refractive index $n_0=1$) is given by: $$\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right),$$ where:

• $n=1.5$ (refractive index of the lens),
• $R_1$ and $R_2$ are the radii of curvature of the lens surfaces.

Step 2: Express the lensmaker’s equation in water

When the lens is immersed in water (refractive index $n_m=1.33$), the relative refractive index of the lens with respect to water becomes: $$n_{\text{relative}}=\frac{n}{n_m}=\frac{1.5}{1.33}\approx \mathrm{1.1278.}$$ The new focal length $f_w$ in water is: $$\frac{1}{f_w}=\left(\frac{n}{n_m}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right).$$

Step 3: Relate the focal lengths

From the lensmaker’s equation in air and water, we derive: $$\frac{1}{f_w}=\left(\frac{n}{n_m}-1\right)\frac{1}{n-1}\cdot \frac{1}{f}.$$ Substituting the known values: $$\frac{1}{f_w}=\left(\frac{1.5}{1.33}-1\right)\frac{1}{1.5-1}\cdot \frac{1}{24}.$$ $$\frac{1}{f_w}=(1.1278-1)\cdot \frac{1}{0.5}\cdot \frac{1}{24}.$$ $$\frac{1}{f_w}=0.1278\cdot 2\cdot \frac{1}{24}=\mathrm{0.01065.}$$

Step 4: Calculate the new focal length

$$f_w=\frac{1}{0.01065}\approx 93.9\mathrm{cm}.$$

Thus, the focal length of the lens in water is 93.9 cm. The focal length increases because the relative refractive index decreases when the lens is immersed in water.