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: Fix the actual radius of curvature first.
The lens is plano-convex, so one face is flat. Taking the curved face with radius \(R\) and the flat face with \(R_2 = \infty\), the geometry factor is \(\dfrac{1}{R_1}-\dfrac{1}{R_2}=\dfrac{1}{R}\).

Step 2: Use the air data to get \(R\).
In air, \(\dfrac{1}{f}=(n_g-1)\dfrac{1}{R}\), so \(\dfrac{1}{10}=\left(\dfrac{3}{2}-1\right)\dfrac{1}{R}=\dfrac{1}{2R}\). Hence \(R = 5\) cm. The lens is a definite piece of glass with a 5 cm curved face.

Step 3: Effective index inside carbon disulphide.
Refraction depends only on the ratio of the two media. Inside carbon disulphide the effective index of the glass is \(\mu' = \dfrac{n_g}{n_c}=\dfrac{3/2}{5/3}=0.9\), which is less than 1. An index less than 1 means light bends the opposite way at the surfaces.

Step 4: Recompute the focal length.
\(\dfrac{1}{f'}=(\mu'-1)\dfrac{1}{R}=(0.9-1)\dfrac{1}{5}=\dfrac{-0.1}{5}=-0.02\ \text{cm}^{-1}\). Therefore \(f' = \dfrac{1}{-0.02} = -50\) cm.

Step 5: Interpret the sign.
Because \(\mu' < 1\), the once-converging lens turns into a diverging lens, and its focal length grows five times in magnitude to 50 cm.

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