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

Step 1: Lens Maker's Formula
The formula is:

$\frac{1}{f} = (\frac{n_2}{n_1} - 1) (\frac{1}{R_1} - \frac{1}{R_2})$

where:

- $f$ is the focal length of the lens
- $n_2$ is the refractive index of the lens material
- $n_1$ is the refractive index of the surrounding medium
- $R_1$ and $R_2$ are the radii of curvature of the lens surfaces

Step 2: Apply Lens Maker's Formula for lens in air

For a glass lens in air, let $f_a$ be the focal length.

The refractive index of glass with respect to air is given as $n_{ag} = \frac{3}{2}$.

Here, $n_2 = n_g = \frac{3}{2}$ and $n_1 = n_{air} = 1$ (approximately).

$\frac{1}{f_a} = (\frac{n_g}{n_{air}} - 1) (\frac{1}{R_1} - \frac{1}{R_2}) = (\frac{3/2}{1} - 1) (\frac{1}{R_1} - \frac{1}{R_2}) = (\frac{3}{2} - 1) (\frac{1}{R_1} - \frac{1}{R_2}) = \frac{1}{2} (\frac{1}{R_1} - \frac{1}{R_2})$

Step 3: Apply Lens Maker's Formula for lens in water

For a glass lens in water, let $f_w$ be the focal length.

The refractive index of glass with respect to water is needed. We are given the refractive index of air to glass ($n_{ag} = \frac{3}{2}$) and air to water ($n_{aw} = \frac{4}{3}$). The refractive index of glass with respect to water ($n_{gw}$) is given by:

$n_{gw} = \frac{n_{ag}}{n_{aw}} = \frac{3/2}{4/3} = \frac{3}{2} \times \frac{3}{4} = \frac{9}{8}$

Here, $n_2 = n_g = \frac{3}{2}$ and $n_1 = n_w = \frac{4}{3}$, or directly use $n_{gw} = \frac{9}{8}$ as the refractive index of lens material with respect to surrounding medium (water).

$\frac{1}{f_w} = (\frac{n_g}{n_w} - 1) (\frac{1}{R_1} - \frac{1}{R_2}) = (\frac{9}{8} - 1) (\frac{1}{R_1} - \frac{1}{R_2}) = (\frac{9-8}{8}) (\frac{1}{R_1} - \frac{1}{R_2}) = \frac{1}{8} (\frac{1}{R_1} - \frac{1}{R_2})$

Step 4: Find the ratio of focal lengths $\frac{f_w}{f_a}$

Divide the equation for $\frac{1}{f_a}$ by the equation for $\frac{1}{f_w}$:

$\frac{1/f_a}{1/f_w} = \frac{\frac{1}{2} (\frac{1}{R_1} - \frac{1}{R_2})}{\frac{1}{8} (\frac{1}{R_1} - \frac{1}{R_2})}$

$\frac{f_w}{f_a} = \frac{1/2}{1/8} = \frac{1}{2} \times \frac{8}{1} = \frac{8}{2} = 4$

Therefore, the ratio of focal lengths of a glass lens in water and in air is $4 : 1$.

Final Answer: The final answer is ${4 : 1}$

Answer 2

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: \[ \frac{f_w}{f_a} = \frac{4}{1} \]