Question

An air bi-convex lens of 10 cm radius of curvature is placed in a cylinder of glass (\(n = \frac{3}{2}\)), as shown in the figure. Find the focal length and nature of the lens. If a liquid of refractive index \(n' = 2\) is filled in the lens, then what will be the power and nature of the lens?
\includegraphics[width=0.5\linewidth]{image9.png}

Hint:

When the lens is filled with a liquid, its focal length decreases, and the power increases. A higher refractive index leads to a stronger lens.

Answer 1

A bi-convex lens is a type of lens in which both surfaces are convex. The focal length of a lens is determined by the lensmaker's formula:
\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] Where: - \(f\) is the focal length of the lens,
- \(n\) is the refractive index of the material of the lens,
- \(R_1\) and \(R_2\) are the radii of curvature of the two surfaces of the lens.
For a bi-convex lens, \(R_1 = +10~\text{cm}\) (convex surface) and \(R_2 = -10~\text{cm}\) (concave surface). The refractive index of the glass is \(n = \frac{3}{2}\).
Substituting these values into the lensmaker’s formula:
\[ \frac{1}{f} = \left( \frac{3}{2} - 1 \right) \left( \frac{1}{10} - \frac{1}{-10} \right) \] \[ \frac{1}{f} = \frac{1}{2} \times \left( \frac{2}{10} \right) \] \[ \frac{1}{f} = \frac{1}{10} \] Thus, the focal length of the lens is:
\[ f = 10~\text{cm} \] The lens is a converging lens because it is bi-convex.
Now, when the lens is filled with a liquid of refractive index \(n' = 2\), we use the modified lensmaker's formula:
\[ \frac{1}{f'} = (n' - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] Where \(n' = 2\) is the refractive index of the liquid. Substituting the values into the formula:
\[ \frac{1}{f'} = \left( 2 - 1 \right) \left( \frac{1}{10} - \frac{1}{-10} \right) \] \[ \frac{1}{f'} = 1 \times \left( \frac{2}{10} \right) \] \[ \frac{1}{f'} = \frac{2}{10} \] Thus, the new focal length of the lens when filled with liquid is:
\[ f' = 5~\text{cm} \] The nature of the lens remains converging, but its focal length decreases due to the increased refractive index.
The power \(P\) of the lens is given by:
\[ P = \frac{1}{f} \] The power in air is:
\[ P = \frac{1}{10} = 0.1~\text{diopters} \] The power in the liquid is:
\[ P' = \frac{1}{5} = 0.2~\text{diopters} \]