A convex lens made of glass (refractive index = 1.5) has focal length 24 cm in air. When it is totally immersed in water (refractive index = 1.33), its focal length changes to

Let's carefully rework the solution.

Step 1: Lens Maker’s Formula

The lens maker’s formula in air is:

$$\frac{1}{f_{\text{air}}} = \left( n_{\text{lens}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$

For a lens immersed in a medium of refractive index $n_{\text{medium}}$, the formula becomes:

$$\frac{1}{f_{\text{medium}}} = \left( \frac{n_{\text{lens}}}{n_{\text{medium}}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$

Dividing these two equations:

$$\frac{f_{\text{medium}}}{f_{\text{air}}} = \frac{n_{\text{lens}} - 1}{\frac{n_{\text{lens}}}{n_{\text{medium}}} - 1}$$

Step 2: Substituting Values

Given:

• $f_{\text{air}} = 24$ cm,
• $n_{\text{lens}} = 1.5$,
• $n_{\text{medium}} = 1.33$ (water).

Now, compute:

$$\frac{f_{\text{water}}}{24} = \frac{1.5 - 1}{\frac{1.5}{1.33} - 1}$$

 $$= \frac{0.5}{\frac{1.5}{1.33} - 1}$$ $$= \frac{0.5}{\frac{1.5 - 1.33}{1.33}}$$ $$= \frac{0.5}{\frac{0.17}{1.33}}$$ $$= \frac{0.5 \times 1.33}{0.17}$$ $$= \frac{0.665}{0.17}$$ $$\approx 3.91$$

$$f_{\text{water}} = 24 \times 3.91$$ $\text{=93.8 cm}\approx 94\text{ cm}$ 

If we use the refractive index of water as $n_{\text{medium}}=1.33=4/3$, 

$\frac{f_{\text{water}}}{24}=\frac{1.5-1}{\frac{1.5\times 3}{4}-1}=4$

$f_{\text{water}}=24\times 4=96 \mathrm{cm}$

Final Answer:

So, when the convex lens is immersed in water, its focal length increases to approximately 96 cm.