Question

An equi convex lens is placed on a horizontal plane mirror. A pin held at a height of 20 cm above the lens concides with its own image. The gap between the lens and the mirror is then filled with water ($\mu = 4 / 3$). To again coincide with its own image, the pin has to be raised by a distance of 7.5 cm.

104. The radius of curvature of surfaces of the lens is:

• A. $\frac{55}{9}cm$
• B. $\frac{100}{9}cm$
• C. $\frac{220}{9}cm$
• D. $\frac{110}{9}cm$

Answer 1

Answer: (C)

$\frac{220}{9}cm$

Answer 2

The problem involves a lens-mirror combination. When a pin coincides with its own image, it means the object is placed at the center of curvature of the equivalent mirror system. Thus, the distance of the pin from the lens is the radius of curvature ($R_{eq}$) of the equivalent mirror, and its focal length is $F_{eq} = R_{eq}/2$. The formula for the focal length of a lens-mirror combination is $\frac{1}{F_{eq}} = \frac{2}{f_L} + \frac{1}{f_M}$. For a plane mirror, $f_M = \infty$, so the formula simplifies to $\frac{1}{F_{eq}} = \frac{2}{f_L}$.

Part 1: Lens in air (initial setup)

The pin is at a height of 20 cm. So, $R_{eq} = 20$ cm. The equivalent focal length is $F_{eq} = 20/2 = 10$ cm. Using the lens-mirror formula: $\frac{1}{10} = \frac{2}{f_L}$ $f_L = 20$ cm. This is the focal length of the equi-convex glass lens in air.

For an equi-convex lens, the radii of curvature are $R_1 = R$ and $R_2 = -R$. Using the lens maker's formula: $\frac{1}{f_L} = (\mu_L - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$ $\frac{1}{20} = (\mu_L - 1) \left( \frac{1}{R} - \frac{1}{-R} \right)$ $\frac{1}{20} = (\mu_L - 1) \left( \frac{2}{R} \right)$ $R = 40 (\mu_L - 1)$ --- (Equation 1)

Part 2: Lens with water in the gap (second setup)

The gap between the lens and the mirror is filled with water ($\mu_w = 4/3$). This setup now consists of two lenses in contact:

1. The original equi-convex glass lens ($L_1$), with focal length $f_1 = f_L = 20$ cm.
2. A plano-concave lens formed by the water layer ($L_2$). Its upper surface is the lower surface of the glass lens (concave for water, radius $R_1 = -R$). Its lower surface is the plane mirror (plane, radius $R_2 = \infty$).

The focal length of the water lens ($f_2$) is: $\frac{1}{f_2} = (\mu_w - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$ $\frac{1}{f_2} = \left( \frac{4}{3} - 1 \right) \left( \frac{1}{-R} - \frac{1}{\infty} \right)$ $\frac{1}{f_2} = \left( \frac{1}{3} \right) \left( \frac{-1}{R} \right)$ $f_2 = -3R$.

The combined focal length of the two lenses ($L_1$ and $L_2$) in contact is $f_{comb}$: $\frac{1}{f_{comb}} = \frac{1}{f_1} + \frac{1}{f_2}$ $\frac{1}{f_{comb}} = \frac{1}{20} + \frac{1}{-3R}$ $\frac{1}{f_{comb}} = \frac{1}{20} - \frac{1}{3R}$

After filling with water, the pin has to be raised by 7.5 cm. New height of the pin = 20 cm + 7.5 cm = 27.5 cm. This new height is the radius of curvature of the new equivalent mirror system, $R'_{eq} = 27.5$ cm. The new equivalent focal length is $F'_{eq} = R'_{eq}/2 = 27.5/2 = 13.75$ cm.

Using the lens-mirror formula for the new system: $\frac{1}{F'_{eq}} = \frac{2}{f_{comb}} + \frac{1}{f_M}$ $\frac{1}{13.75} = \frac{2}{f_{comb}}$ $f_{comb} = 2 \times 13.75 = 27.5$ cm.

Now, equate the two expressions for $f_{comb}$: $\frac{1}{27.5} = \frac{1}{20} - \frac{1}{3R}$ $\frac{1}{3R} = \frac{1}{20} - \frac{1}{27.5}$ $\frac{1}{3R} = \frac{1}{20} - \frac{10}{275}$ $\frac{1}{3R} = \frac{1}{20} - \frac{2}{55}$ To subtract, find a common denominator, which is 220: $\frac{1}{3R} = \frac{11}{220} - \frac{8}{220}$ $\frac{1}{3R} = \frac{3}{220}$ $3R = \frac{220}{3}$ $R = \frac{220}{9}$ cm.

The radius of curvature of surfaces of the lens is $\frac{220}{9}$ cm.