# Michelson Interferometer: Optics

1. Feb 18, 2014

### unscientific

1. The problem statement, all variables and given/known data

A Michelson Interferometer has incident light in range 780-920 nm from a source. Intensity as a function of x (distance from central maxima) is given by:

$$I_{(x)} = 3I_0 + 3I_0 cos(K_1x) cos (K_2x) - I_0 sin (K_1x) sin (K_2x)$$

where $K_1 = 1.3 x 10^7 m^{-1}$ and $K_2 = 3.3 x 10^5 m^{-1}$ and $I_0$ is intensity of central maxima.

Part(a): Show that intensity can be written as sum of two monochromatic components.

Part(b): Find their mean wavenumber $\overline {v}$

Part(c): Find their wavenumber separation $\Delta \overline {v}$

Part (d): Find their relative intensities

Part (e): Over a larger range, it is found that periodic terms are in fact multiplied by:

$$f_{(x)} \propto e^{(-K_3^2 x^2)}$$

Make a sketch of the interference pattern.

2. Relevant equations

3. The attempt at a solution

Part(a)

$$I_{(x)} = 3I_0 + 3I_0 cos(K_1x) cos (K_2x) - I_0 sin (K_1x) sin (K_2x)$$
$$= 3I_0 + 2I_0 cos\left ( (k_1 + k_2) x\right ) + I_0 cos \left ((k_1 - k_2)x \right )$$
$$= \frac{3}{2}I_0 + 2I_0 cos (k_3' x) + \frac{3}{2}I_0 + I_0 cos (k_4' x)$$

where $k_3' = k_1 + k_2$ and $k_4' = k_1 - k_2$.

Thus it is a sum of two monochromatic intensities with wavenumber $k_3'$ and $k_4'$.

Part (b)

$$\overline {v} = \frac{k_3' + k_4'}{2} = k_1$$

Part (c)

$$\Delta \overline {v} = \frac{k_3' - k_4'}{2} = k_2$$

Part(d)

$$\frac{I'_3}{I'_4} = \frac{\frac{3}{2} + 2 cos (k_3' x)}{\frac{3}{2} + cos (k_4' x)}$$

At x = 0,

$$\frac{I'_3}{I'_4} = \frac{7}{4}$$

For relative intensities, do they want us to find it at a distance x away or at the center?

Part (e)

I can see that for large x, Intensity → 6I0

$$I = \frac{3}{2}I_0 + 2I_0 cos (Ak_3' e^{x^2}) + \frac{3}{2}I_0 + I_0 cos (Ak_4' e^{x^2})$$

So the general idea is that it varies sinusoidally about 6I0 but finally settles along 6I0. Here's my sketch: