Question

If a microscope is placed in air, the minimum separation of two objects seen as distinct is 6 µm. If the same is placed in a medium of refractive index 1.5, the minimum separation of the two objects to see as distinct is:

• A. \(4 \, \mu m\)
• B. \(6 \, \mu m\)
• C. \(3 \, \mu m\)
• D. \(9 \, \mu m\)

Hint:

The resolvability of microscopic images improves in media with higher refractive indices because the effective wavelength of light inside the medium decreases, allowing for finer details to be resolved.

Answer 1

The Correct Option is A

The minimum resolvable distance \( d \) in a microscope can be related to the refractive index \( n \) and the wavelength \( \lambda \) using the formula: \[ d = \frac{\lambda}{n} \] Assuming the wavelength in air is the same as the minimum resolvable distance when \( n = 1 \), that is, \( \lambda = 6 \, \mu m \): When the medium changes to a refractive index of \( n = 1.5 \), the new resolvable distance is: \[ d' = \frac{6 \, \mu m}{1.5} = 4 \, \mu m \]