Question

An isolated nucleus from a cell is on a microscope slide and is surrounded by growth medium. The nucleus has a refractive index of 1.42, while the surrounding fluid has an index of 1.35.

For a wavelength of 500 nm, what will be the relative phase shift between light that is transmitted through the center of the nucleus vs. light that goes through the fluid?

Can you use this information to explain why the phase of the unretarded light needs to be shifted by ? /2 (pi/2) to accomplish good phase contrast imaging?

Answer 1

General Theory,

x is the path difference.

Now coming to your question. You won't be needing these formulae. This question is theoretical.

(a) After reflection from a denser medium, the phase is shifted by pi. This does not happen when the reflection occurs when light reflects from a denser medium.

Calculating the reflectance vector of light as follows:

As r is positive, there is no phase change reflection from a rarer medium. A phase change of 180 degrees would have occurred upon reflection from a denser medium. Also, there is no phase change associated with refraction. This is because the Electric flux needs to be equal on either side of the interface.

Therefore,

relative phase shift = 0.

(b) When observing an unstained biological specimen, the scattered light is weak and typically phase-shifted by ?90° (due to both the typical thickness of specimens and the refractive index difference between biological tissue and the surrounding medium) relative to the background light. This leads to the foreground and background having nearly the same intensity, resulting in low image contrast. The background light is phase-shifted by ?90° by passing it through a phase-shift ring, which eliminates the phase difference between the background and the scattered light rays.