Question

A convex lens of refractive index \(15\) and focal length\(18 \,cm\) in air is immersed in water. The change in focal length of the lens will be_______ cm (Given refractive index of water =\(\frac{4}{3}\))

Hint:

The focal length changes when a lens is immersed in a medium. Use \[ \mu_\text{eff} = \frac{\mu_\text{lens}}{\mu_\text{medium}} \] in the lens maker's formula.

Answer 1

Correct Answer: 54

The focal length of a lens in water is given by:

\[ \frac{1}{f_\text{water}} = \left( \frac{\mu_\text{glass}}{\mu_\text{water}} - 1 \right) \frac{2}{R} \]

For air:  \[ \frac{1}{f_\text{air}} = (\mu_\text{glass} - 1) \frac{2}{R} \]

Given:

• \(\mu_\text{glass} = 1.5\),
• \(\mu_\text{water} = \frac{4}{3}\),
• \(f_\text{air} = 18 \, \text{cm}\).

The ratio of focal lengths in water and air is:

\[ \frac{f_\text{water}}{f_\text{air}} = \frac{\frac{1}{f_\text{air}}}{\frac{1}{f_\text{water}}} = \frac{\mu_\text{glass} - 1}{\frac{\mu_\text{glass}}{\mu_\text{water}} - 1} \]

Substitute the values:

\[ \frac{f_\text{water}}{f_\text{air}} = \frac{1.5 - 1}{\frac{1.5}{\frac{4}{3}} - 1} = \frac{0.5}{2 - 1} = \frac{0.5}{1} = 0.5 \]

Thus:

\[ f_\text{water} = 4 f_\text{air} = 4 \times 18 = 72 \, \text{cm} \]

The change in focal length is:

\[ \Delta f = f_\text{water} - f_\text{air} = 72 - 18 = 54 \, \text{cm} \]

Answer 2

The correct answer is 54.
fH2​O​I​=(μH2​O​μg​​−1)(R2​) 
=81​(R2​) 
=(4fair​)1​ 
So, fH2​O​=4fair ​=72cm 
So change in focal length =72−18=54cm