Question

If the refractive index from air to glass is \(\frac{3}{2}\) and that from air to water is \(\frac{4}{3}\), then the ratio of focal lengths of a glass lens in water and in air is

• A. 1 : 2
• B. 2 : 1
• C. 1 : 4
• D. 4 : 1

Answer 1

The Correct Option is D

Focal Length Ratio in Different Media

Given the refractive indices of water (\( n_w \)) and glass (\( n_g \)):

\[ n_w = \frac{4}{3}, \quad n_g = \frac{3}{2} \]

The relationship between the focal lengths in air (\( f_a \)) and water (\( f_w \)) is derived from the lensmaker's equation:

\[ f_a (n_g - 1) = f_w \left( \frac{n_g}{n_w} - 1 \right) \]

Solving for the ratio \( \frac{f_w}{f_a} \):

\[ \begin{align*} \frac{f_w}{f_a} &= \frac{n_g - 1}{\frac{n_g}{n_w} - 1} \\ &= \frac{\frac{3}{2} - 1}{\frac{3/2}{4/3} - 1} \\ &= \frac{\frac{1}{2}}{\frac{9}{8} - 1} \\ &= \frac{\frac{1}{2}}{\frac{1}{8}} \\ &= 4 \end{align*} \]

Therefore, the ratio of the focal lengths is: \[ \frac{f_w}{f_a} = \frac{4}{1} \]