Question

The focal length of a glass \((n_g = 3/2)\) plano-convex lens in air is 10 cm. What will be the focal length and its nature when it is immersed in carbon disulphide \((n_c = 5/3)\)?

Hint:

Use \(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\) with \(\mu\) taken relative to the medium. In carbon disulphide \(\mu=n_g/n_c=9/10<1\), so the lens turns diverging.

Answer 1

Step 1: Write the lens maker's formula.
\[ \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
where \(\mu\) is the refractive index of the lens material with respect to the surrounding medium.

Step 2: Apply it in air.
In air the surrounding index is 1, so \(\mu = n_g = \dfrac{3}{2}\) and \(f = 10\) cm.
\[ \frac{1}{10} = \left(\frac{3}{2} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{2}\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
Therefore the fixed geometry factor is
\[ \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{2}{10} = \frac{1}{5}\ \text{cm}^{-1} \]

Step 3: Find the new relative refractive index in carbon disulphide.
Now the lens is surrounded by carbon disulphide, so
\[ \mu' = \frac{n_g}{n_c} = \frac{3/2}{5/3} = \frac{3}{2}\times\frac{3}{5} = \frac{9}{10} \]

Step 4: Apply the lens maker's formula in carbon disulphide.
The geometry factor \(\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\) does not change, so
\[ \frac{1}{f'} = (\mu' - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \left(\frac{9}{10} - 1\right)\times\frac{1}{5} \]
\[ \frac{1}{f'} = \left(-\frac{1}{10}\right)\times\frac{1}{5} = -\frac{1}{50} \]

Step 5: Result and nature.
\[ f' = -50\ \text{cm} \]
The magnitude of the focal length increases from 10 cm to 50 cm, and the negative sign shows the lens now behaves as a diverging (concave-acting) lens. This happens because carbon disulphide is optically denser than glass, so the converging lens loses its converging power and becomes diverging.

\[\boxed{f' = -50\ \text{cm, diverging lens}}\]