Light of wavelength 589 nm is used to view an object under a microscope. The aperature of the objective has a diameter of `0.900 cm`. Find
What effect would this have on the resolving power, if water `(mu = 1.33)` fills the space between the object and objective.

Explanation

To solve the problem, we need to determine how filling the space between the object and the objective lens with water affects the resolving power of the microscope. The resolving power \( R \) of a microscope can be expressed using the formula:

\[ R = \frac{2 \mu \sin \theta}{\lambda} \]

Where:
- \( R \) is the resolving power.
...