# Michelson Interferometer - calculating errors

1. Apr 20, 2014

### unscientific

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

Part (a): Find an expression for intensity, interference phase and total change of interference phase.

Part (b): Find an expression for <L1>/L0 and its error.

Part (c): Find an expression for Δø1 and proportionality constant.

Part (d): Find an expression for n(0) and α.

2. Relevant equations

3. The attempt at a solution

Part (a)

$$v_t = (<v> - \frac{\Delta v}{2}) + \Delta v(\frac{t}{T})$$

The phase difference is given by $\Phi_0 = \frac{2\pi}{\lambda} (2L_0) = 4\pi v_t L_0$.

$$\Phi_0 = 4\pi L_0 \left[ <v> + \frac{\Delta v}{2}(\frac{2t}{T} - 1)\right]$$

Change in phase:
$$\Delta \Phi_0 = \Phi_{0(T)} - \Phi_{0(0)} = 4\pi L_0 (\Delta v)$$

Part (b)
$$\Delta \Phi_0 = 4\pi L_0 (\Delta v)$$
$$\Delta \Phi_1 = 4\pi L_1 (\Delta v)$$

Therefore,$\frac{<L_1>}{L_0} = \frac{\Delta \Phi_1}{\Delta \Phi_0} = \frac{\sigma_{(\Delta \Phi_1})}{\sigma_{(\Delta \Phi_0)}}$

To find the change of this, consider $x = \frac{\Delta y}{\Delta z}$, $dx = (\frac{\partial x}{\partial y})dy + (\frac{\partial x}{\partial z})dz$. Therefore $\sigma_{(\frac{<L_1>}{L_0})} = 2 \frac{\sigma_{(\Delta \Phi_1)}}{\sigma_{(\Delta \Phi_0)}}$.

Part (c)

Now the length is replaced by:
$$L_{1(t)} = <L_1> + \frac{\Delta L_1}{2}\left[ 2\frac{t}{T} - 1\right]$$
$$= 4\pi(<L_1> + \frac{\Delta L_1}{2})(<v> + \frac{\Delta v}{2}) - 4\pi(<L_1> - \frac{\Delta L_1}{2})(<v> - \frac{\Delta v}{2})$$
$$= 4\pi\left[\Delta v <L_1> + \Delta L_1 <v>\right]$$

The proportionality constant $C = \frac{\sigma_{\Delta L_1}}{\Delta L_1}$ I have no idea how to find it.

Part (d)

Using $n_i = n(<v> - \frac{\Delta v}{2})$ and $n_f = n(<v> + \frac{\Delta v}{2})$, and since $v_t$ grows linearly with time, therefore $n_{(\gamma)}$ grows linearly with time.

Thus $n_{(\gamma)} = n_i + \frac{(n_f - n_i)}{T}t = n_i + \Delta v(\frac{t}{T})$

Equating to the expression given: $n_{(\gamma)} = n_{(0)} + \alpha \gamma = n_0 + \alpha \left[<v> + \frac{\Delta v}{2}(\frac{2t}{T} - 1)\right]$:

$$\alpha = \frac{n_i - n_0}{(<v> - \frac{\Delta v}{2})}$$

To solve for $n_{(0)}$, we need path difference to balance out:
$$\sigma_{L_1}= C\Delta L_1 = \frac{L_b}{ n_{(0)} + \alpha \gamma = n_0 + \alpha \left[<v> + \frac{\Delta v}{2}(\frac{2t}{T} - 1)\right]}$$

But this seems strange as there is a time dependence in the denominator.. Also, how do I solve for proportionality constant C in part (c)?

2. Apr 21, 2014

### unscientific

bumpp

3. Apr 22, 2014

### unscientific

Part (c) - how do i find the proportionality constant?
Part (d) - it is weird that I end up with a time dependence in the error of L1