Question

An equi-convex lens of refractive index (3/2) and focal length 10 cm is held with its axis vertical and its lower surface immersed in water ($\mu$ = 4/3), the upper surface being in air. At what distance from the lens, will a vertical beam of parallel light incident on the lens be focused?

Answer 1

20 cm

Answer 2

The problem asks us to find the focal point of a parallel beam of light passing through an equi-convex lens, where one side is in air and the other is in water.

1. Determine the radius of curvature (R) of the lens:

An equi-convex lens has radii of curvature of equal magnitude. Let the magnitude be $R$. For light incident from the left, the first surface is convex ($R_1 = +R$) and the second surface is also convex, but its center of curvature is on the incident side ($R_2 = -R$).

The focal length of the lens in air ($f_{air} = 10$ cm) is given. The refractive index of the lens is $\mu_L = 3/2$.

Using the lens maker's formula:

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

$\frac{1}{10} = \left(\frac{3}{2} - 1\right) \left(\frac{1}{R} - \frac{1}{-R}\right)$

$\frac{1}{10} = \left(\frac{1}{2}\right) \left(\frac{2}{R}\right)$

$\frac{1}{10} = \frac{1}{R}$

So, $R = 10$ cm.

2. Calculate the power of each surface:

The light travels from air ($\mu_{air} = 1$) into the lens ($\mu_L = 3/2$) through the first surface, and then from the lens ($\mu_L = 3/2$) into water ($\mu_W = 4/3$) through the second surface.

The power of a spherical refracting surface is given by $P = \frac{\mu_2 - \mu_1}{R}$.

• Power of the first surface (air-lens interface):

  $\mu_1 = \mu_{air} = 1$, $\mu_2 = \mu_L = 3/2$, $R_1 = +10$ cm.

  $P_1 = \frac{\mu_L - \mu_{air}}{R_1} = \frac{3/2 - 1}{+10} = \frac{1/2}{10} = \frac{1}{20} \text{ cm}^{-1}$

• Power of the second surface (lens-water interface):

  $\mu_1 = \mu_L = 3/2$, $\mu_2 = \mu_W = 4/3$, $R_2 = -10$ cm (as its center of curvature is on the left side, opposite to the direction of light propagation).

  $P_2 = \frac{\mu_W - \mu_L}{R_2} = \frac{4/3 - 3/2}{-10} = \frac{8/6 - 9/6}{-10} = \frac{-1/6}{-10} = \frac{1}{60} \text{ cm}^{-1}$

3. Calculate the total power of the lens system:

For a thin lens, the total power is the sum of the powers of its individual surfaces:

$P_{total} = P_1 + P_2 = \frac{1}{20} + \frac{1}{60} = \frac{3 + 1}{60} = \frac{4}{60} = \frac{1}{15} \text{ cm}^{-1}$

4. Find the image position (focal length) for a parallel beam:

A parallel beam of light means the object is at infinity ($u = -\infty$). The light originates in air ($\mu_{initial} = \mu_{air} = 1$) and finally focuses in water ($\mu_{final} = \mu_W = 4/3$).

The general formula relating power, object distance, and image distance for a system with different media on either side is:

$\frac{\mu_{final}}{v} - \frac{\mu_{initial}}{u} = P_{total}$

$\frac{4/3}{v} - \frac{1}{-\infty} = \frac{1}{15}$

$\frac{4}{3v} - 0 = \frac{1}{15}$

$\frac{4}{3v} = \frac{1}{15}$

$3v = 4 \times 15$

$3v = 60$

$v = 20$ cm

The positive sign for $v$ indicates that the image is formed on the right side of the lens, in the water.