Question

An objective lens with half angular aperture 𝛼 is illuminated with light of wavelength 𝜆. The refractive index of the medium between the sample and the objective is 𝑛. The lateral resolving power of the optical system can be increased by

• A. decreasing both 𝜆 and α
• B. decreasing 𝜆 and increasing α
• C. increasing both 𝛼 and n
• D. ecreasing 𝜆 and increasing n

Answer 1

The Correct Option is B, C, D

The concept of lateral resolving power of an optical system is tied to its ability to distinguish small details of the sample. The lateral resolution \( R \) is defined by the formula:

\(R = \frac{0.61 \lambda}{n \sin \alpha}\)

where: 

• \(\lambda\) is the wavelength of light used.
• \(n\) is the refractive index of the medium between the sample and the objective.
• \(\alpha\) is the half angular aperture of the objective lens.

To increase the resolving power (or decrease the resolution value \( R \)), we need to:

1. Decrease the wavelength \(\lambda\), as it is directly proportional to \( R \).
2. Increase the refractive index \(n\), as it appears in the denominator of the resolution formula, thus reducing \( R \).
3. Increase the half angular aperture \(\alpha\), which is part of the \( \sin \alpha \) term in the denominator, again reducing \( R \).

Given these insights, the options increasing the resolving power are:

• Decreasing \(\lambda\) and increasing \(\alpha\).
• Increasing both \(\alpha\) and \(n\).
• Decreasing \(\lambda\) and increasing \(n\).

Thus, the options that correctly increase the lateral resolving power, as mentioned, are:

• Decreasing \(\lambda\) and increasing \(\alpha\)
• Increasing both \(\alpha\) and \(n\)
• Decreasing \(\lambda\) and increasing \(n\)

These measures optimize the resolving power of the optical system, making it capable of differentiating finer details in illuminated samples.

Diagram illustrating the half angular aperture \( \alpha \).