Understanding Refractive Index and Focal Length of a Lens

The focal length of a lens is not a fixed property; it depends on the refractive index of the material of the lens and the refractive index of the surrounding medium. When a lens is placed in a medium other than air, its focal length changes because the relative refractive index between the lens material and the surrounding medium changes.

Applying the Lens Maker's Formula

The relationship between the focal length (\(f\)) of a lens, its radii of curvature (\(R_1\) and \(R_2\)), and the refractive index (\(\mu\)) of the lens material relative to the surrounding medium is given by the Lens Maker's Formula:

\(\frac{1}{f} = (\mu_{rel} - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\)

Here, \(\mu_{rel} = \frac{\mu_{lens}}{\mu_{medium}}\) is the refractive index of the lens material relative to the medium it is placed in. The term \( \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \) depends only on the shape of the lens and remains constant.

Focal Length in Air and Water

Let \(f_a\) be the focal length of the lens in air and \(f_w\) be the focal length of the lens in water. Let \(\mu_g\) be the refractive index of the glass (lens material) and \(\mu_w\) be the refractive index of water, both relative to air (\(_a\mu_g\) and \(_a\mu_w\)). The refractive index of air relative to itself is approximately 1.

For the lens in air, the Lens Maker's formula is:

\(\frac{1}{f_a} = \left(\frac{\mu_g}{\mu_{air}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (\mu_g - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\) (Since \(\mu_{air} \approx 1\))

For the lens in water, the formula is:

\(\frac{1}{f_w} = \left(\frac{\mu_g}{\mu_w} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\)

Calculating the Ratio of Focal Lengths

To find the focal length in water (\(f_w\)), we can divide the two equations:

\(\frac{1/f_w}{1/f_a} = \frac{\left(\frac{\mu_g}{\mu_w} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)}{(\mu_g - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)}\)

The term \( \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \) cancels out, leaving:

\(\frac{f_a}{f_w} = \frac{\left(\frac{\mu_g}{\mu_w} - 1\right)}{(\mu_g - 1)}\)

Substituting Given Refractive Indices and Values

We are given:

• Refractive index of water w.r.t. air, \(_a\mu_w = \mu_w = \frac{4}{3}\)
• Refractive index of glass w.r.t. air, \(_a\mu_g = \mu_g = \frac{3}{2}\)
• Focal length in air, \(f_a = 10\) cm

We need to find the refractive index of glass w.r.t. water, \(\frac{\mu_g}{\mu_w}\).

\(\frac{\mu_g}{\mu_w} = \frac{_a\mu_g}{_a\mu_w} = \frac{3/2}{4/3} = \frac{3}{2} \times \frac{3}{4} = \frac{9}{8}\)

Now substitute the values into the ratio equation:

\(\frac{10}{f_w} = \frac{\left(\frac{9}{8} - 1\right)}{\left(\frac{3}{2} - 1\right)}\)

Calculate the terms in the numerator and denominator:

• Numerator: \(\frac{9}{8} - 1 = \frac{9}{8} - \frac{8}{8} = \frac{1}{8}\)
• Denominator: \(\frac{3}{2} - 1 = \frac{3}{2} - \frac{2}{2} = \frac{1}{2}\)

So, the equation becomes:

\(\frac{10}{f_w} = \frac{1/8}{1/2} = \frac{1}{8} \times \frac{2}{1} = \frac{2}{8} = \frac{1}{4}\)

Solving for Focal Length in Water

We have the equation:

\(\frac{10}{f_w} = \frac{1}{4}\)

Cross-multiplying gives:

\(f_w = 10 \times 4\)

\(f_w = 40\) cm

Thus, the focal length of the lens when placed inside water is 40 cm. This demonstrates how the surrounding medium's refractive index significantly affects the lens's focal length compared to when it's in air. The change in refractive index leads to a longer focal length in water.