Mach-Zehnder Interferometer

Step 1/16
a) What is the output power of the interferometer?

Step 2/16
Understand the power splitting ratios of the beamsplitters. Beamsplitter A has a transmission:reflection ratio of 20:80, meaning 20% of the power is transmitted and 80% is reflected. Beamsplitter B has a 50:50 ratio, equally splitting the power.

Step 3/16
Calculate the power after Beamsplitter A. With an input power of 30 mW and a 20:80 split, the transmitted power through Beamsplitter A is $$ (30 mW \times 0.2 = 6 mW) $$, and the reflected power is $$ (30 mW \times 0.8 = 24 mW) $$.

Step 4/16
Since the reflected power does not contribute to the output in this setup, focus on the transmitted power. The 6 mW transmitted through Beamsplitter A then encounters Beamsplitter B.

Step 5/16
Calculate the power after Beamsplitter B. With a 50:50 split, the 6 mW is divided equally into two paths, each receiving $$ (6 mW \times 0.5 = 3 mW)$$.

Step 6/16
Since both paths after Beamsplitter B contribute to the output, the total output power is the sum of the power from both paths, which is $$(3 mW + 3 mW = 6 mW)$$. b) Plot the power variation as a function of the translation stage displacement $$((l))$$, within a range of $$(l = 0) to (l = 5 \mu m)$$.

Step 7/16
Understand that the power variation due to the path length difference is governed by the interference pattern, which depends on the phase difference between the two paths.

Step 8/16
The phase difference$$ ((\Delta \phi))$$ is related to the path length difference$$ ((\Delta l)) and the wavelength ((\lambda)) $$by the equation $$(\Delta \phi = \frac{2\pi \Delta l}{\lambda})$$.

Step 9/16
The intensity (or power) variation due to interference can be described by$$ (I = I_0 \cos^2(\frac{\Delta \phi}{2}))$$, where $$(I_0)$$ is the maximum intensity (3 mW for each path in this case).

Step 10/16
To plot this, you would calculate$$ (I) $$for values of$$ (l) $$ranging from 0 to$$ (5 \mu m) $$and plot$$ (I) $$versus$$ (l)$$. This requires computational tools like MATLAB or Python. The plot will show oscillations due to the cosine squared term, reflecting constructive and destructive interference as $$(l)$$ changes. c) Power variation with an LED source of bandwidth$$ (v = 35 THz)$$.

Step 11/16
Understand that a larger bandwidth means the source has a wider range of wavelengths. This leads to a decrease in coherence length, which is the length over which the light waves maintain a fixed phase relationship.

Step 12/16
With a coherence length much shorter than the path length differences explored in the experiment, the interference fringes will wash out, leading to a reduced visibility of the power variation as a function of $$(l)$$.

Step 13/16
The reason for this is that different wavelengths interfere at different rates, and with a wide range of wavelengths, these interferences will not be in sync, leading to an averaging effect that smooths out the variations. d) Modifying the setup for the LED source.

Step 14/16
To observe power variation with the LED source, one needs to increase the coherence length of the light entering the interferometer. This can be achieved by narrowing the bandwidth of the light.

Step 15/16
Use a monochromator or a narrow bandpass filter before the interferometer to select a narrow range of wavelengths from the LED source. This will effectively increase the coherence length of the light entering the interferometer.

Answer
Ensure that the selected bandwidth is narrow enough to provide a coherence length longer than the maximum path length difference explored in the experiment. This will allow the observation of interference patterns and thus the power variation with the translation stage displacement.