Question

A convex lens of glass of focal length 15 cm is immersed in carbon disulphide. In this situation what will be the focal length and the nature of the lens? The refractive indices of glass and carbon disulphide (relative to air) are 3/2 and 5/3 respectively.

Hint:

Apply the lens maker's formula in both media; the shape factor is common, so f_medium/f_air = (n_g - 1)/((n_g/n_CS2) - 1). Since carbon disulphide is denser than glass, the convex lens turns diverging.

Answer 1

Step 1: Lens maker's formula.
For a lens of refractive index \( n_{lens} \) placed in a medium of refractive index \( n_{med} \),
\[ \frac{1}{f} = \left(\frac{n_{lens}}{n_{med}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
The shape factor \( \left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right) \) depends only on the geometry of the lens and stays the same in any medium.

Step 2: In air.
Here \( n_{med} = 1 \), glass \( n_g = \dfrac{3}{2} \) and \( f_{air} = 15\ \text{cm} \).
\[ \frac{1}{15} = \left(\frac{3}{2} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{2}\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
\[ \therefore \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{2}{15}\ \text{cm}^{-1} \]

Step 3: In carbon disulphide.
Refractive index of glass relative to carbon disulphide:
\[ \frac{n_g}{n_{CS_2}} = \frac{3/2}{5/3} = \frac{3}{2}\times\frac{3}{5} = \frac{9}{10} \]
So
\[ \frac{1}{f_{med}} = \left(\frac{9}{10} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \left(-\frac{1}{10}\right)\times\frac{2}{15} \]

Step 4: Arithmetic.
\[ \frac{1}{f_{med}} = -\frac{2}{150} = -\frac{1}{75}\ \text{cm}^{-1} \]
\[ f_{med} = -75\ \text{cm} \]

Step 5: Interpretation.
The focal length becomes \( -75\ \text{cm} \). The negative sign shows that the converging (convex) glass lens now behaves as a diverging (concave) lens. This happens because carbon disulphide (\( n = 5/3 \)) is optically denser than glass (\( n = 3/2 \)).
\[\boxed{f = -75\ \text{cm};\ \text{it acts as a diverging (concave) lens}}\]