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: Use the ratio of the two lens equations.
The geometrical factor \(\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\) is common to the lens in air and in the liquid, so dividing the two lens maker equations eliminates it:
\[\frac{f_{med}}{f_{air}} = \frac{\left(\dfrac{n_g}{n_{air}} - 1\right)}{\left(\dfrac{n_g}{n_{med}} - 1\right)}\]

Step 2: Insert the refractive indices.
Glass in air: \(\frac{n_g}{n_{air}} - 1 = \frac{3}{2} - 1 = \frac{1}{2}\).
Glass in carbon disulphide: \(\frac{n_g}{n_{med}} - 1 = \frac{3/2}{5/3} - 1 = \frac{9}{10} - 1 = -\frac{1}{10}\).

Step 3: Form the ratio.
\[\frac{f_{med}}{f_{air}} = \frac{1/2}{-1/10} = \frac{1}{2}\times(-10) = -5\]

Step 4: Solve for the new focal length.
\[f_{med} = -5 \times f_{air} = -5 \times 15 = -75\ \text{cm}\]

Step 5: Nature of the lens.
The magnitude of the focal length increases from 15 cm to 75 cm and the sign turns negative. A negative focal length means the convex lens now diverges light, so in carbon disulphide it acts as a concave (diverging) lens, because the surrounding liquid is optically denser than the glass.
\[\boxed{f = -75\ \text{cm};\ \text{diverging (concave) behaviour}}\]