Question

A spherical air lens of radii \( R_1 = R_2 = 10 \, \text{cm} \) is cut from a glass (\(\mu = 1.5\)) cylinder as shown in the figure. Its focal length is \( f_1 \). If a liquid of refractive index \( \mu_2 \) is filled in the space, then the focal length of the liquid lens becomes \( f_2 \). Calculate \( f_1 \) and \( f_2 \). Choose the correct options from the following.

• A. \( f_1 = 15 \, \text{cm}, f_2 = 30 \, \text{cm} \)
• B. \( f_1 = -15 \, \text{cm}, f_2 = +30 \, \text{cm} \)
• C. \( f_1 = -15 \, \text{cm}, f_2 = +15 \, \text{cm} \)
• D. \( f_1 = -30 \, \text{cm}, f_2 = -15 \, \text{cm} \)

Hint:

The focal length of a lens is inversely proportional to the radii of curvature and the refractive index difference. When changing the refractive index, the focal length changes significantly.

Answer 1

The Correct Option is A

For a spherical lens, the focal length \( f \) is given by the lens formula: \[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] For the air lens: \[ \frac{1}{f_1} = (\mu_{\text{glass}} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = (1.5 - 1) \left( \frac{1}{10} - \frac{1}{10} \right) \] \[ \frac{1}{f_1} = 0.5 \times \left( \frac{2}{10} \right) = \frac{0.1}{10} \quad \Rightarrow \quad f_1 = 15 \, \text{cm} \] When a liquid with refractive index \( \mu_2 \) is filled, the effective refractive index becomes: \[ \frac{1}{f_2} = (\mu_2 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] For \( \mu_2 = 1.5 \), the focal length is: \[ \frac{1}{f_2} = (1.5 - 1) \left( \frac{2}{10} \right) = \frac{1}{5} \quad \Rightarrow \quad f_2 = 30 \, \text{cm} \]