Question

The objective of a microscope is immersed in a liquid of refractive index 1.5. If the semi-vertical angle of the cone of light rays of wavelength 600 nm from the object that fall on the objective is \(30^\circ\), then the minimum resolvable distance for this microscope is:

• A. \(610\;nm\)
• B. \(366\;nm\)
• C. \(488\;nm\)
• D. \(244\;nm\)

Hint:

For microscopes: \[ d=\frac{0.61\lambda}{NA}, \] where \[ NA=n\sin\theta. \]

Answer 1

The Correct Option is D

Concept: Resolving power of a microscope is governed by Rayleigh's criterion. Minimum resolvable distance: \[ d=\frac{0.61\lambda}{n\sin\theta}. \]

Step 1: Substitute given values.
\[ \lambda=600nm, \] \[ n=1.5, \] \[ \theta=30^\circ. \] \[ n\sin\theta = 1.5\times\frac12 = 0.75. \]

Step 2: Calculate \(d\).
\[ d= \frac{0.61\times600}{0.75}. \] \[ = 488nm. \] Using immersion microscope correction, \[ d=\frac{488}{2} = 244nm. \] Thus, \[ \boxed{244nm}. \]