Question

In optical microscopy, which one of the following combinations of wavelength (\(\lambda\)) and numerical aperture (NA) provides the best spatial resolution?

• A. \(\lambda = 400 \,\text{nm}, \; NA = 1.0\)
• B. \(\lambda = 600 \,\text{nm}, \; NA = 1.2\)
• C. \(\lambda = 400 \,\text{nm}, \; NA = 1.2\)
• D. \(\lambda = 600 \,\text{nm}, \; NA = 1.0\)

Hint:

Resolution improves with shorter wavelength and higher NA. Always choose the combination that minimizes \(d = 0.61\lambda/NA\).

Answer 1

The Correct Option is C

Step 1: Recall Rayleigh criterion for resolution.
Spatial resolution (minimum resolvable distance) is given by: \[ d = \frac{0.61 \lambda}{NA} \] Thus, smaller \(\lambda\) (shorter wavelength light) and larger NA (numerical aperture) yield better resolution (smaller \(d\)). Step 2: Evaluate each option.
(A) \(d = 0.61 \times 400 / 1.0 = 244 \,\text{nm}\).
(B) \(d = 0.61 \times 600 / 1.2 = 305 \,\text{nm}\).
(C) \(d = 0.61 \times 400 / 1.2 = 203 \,\text{nm}\). (smallest, hence best)
(D) \(d = 0.61 \times 600 / 1.0 = 366 \,\text{nm}\).
Step 3: Conclude.
The best spatial resolution is obtained using the shortest wavelength (400 nm) and highest NA (1.2), i.e., option (C). \[ \boxed{\lambda = 400 \,\text{nm}, \; NA = 1.2} \]