Question

If the angular aperture of a 20X objective is 46°, the numerical aperture of the water immersion objective is_____. (Round off to two decimal places) (Use RI of water = 1.33)

Answer 1

Correct Answer: 0.5

Given:
Angular aperture of objective: \(\theta = 46^\circ\)
Refractive index of water: \(n = 1.33\)

Step 1 — Recall the formula for numerical aperture (NA): 
\[ \text{NA} = n \cdot \sin\left(\frac{\theta}{2}\right) \] 
Step 2 — Plug in the values:
\[ \text{NA} = 1.33 \cdot \sin\left(\frac{46^\circ}{2}\right) = 1.33 \cdot \sin 23^\circ \] \[ \sin 23^\circ \approx 0.3907 \] \[ \text{NA} = 1.33 \cdot 0.3907 \approx 0.52 \] Answer (rounded to two decimals): \(\boxed{0.52}\)