Question

04. As noted in the Experiment 4 hand-out, the intensity I that results from superposition of two coherent light beams at the photodiode detector is given by I =i, +12 + 2112 cos ? , 2? where ? = k (sl-s2)-EAs is the phase difference in radians between the two beams due to having traveled different path lengths s, and s2. Defining r 2/, we can rewrite that as 12vT cos 8] When referring to a Michelson interferometer, this r factor reflects some imbalance between the two arms due to an imperfect beam splitter and/or optical losses somewhere. Now, when using a sodium lamp, the two wavelengths that make up the "D line doublet" are produced with somewhat different intensities but should have the same r factor due to the interferometer. Let the irradiance for the light of wavelength ?1 be 11, and that for wavelength ?2 be 1,-q11. Here, 11 and 12 are not the signals for the two arms, but simply the signals associated with the two different wavelengths. The relative strength of the signal:s traveling through the two arms is still r. a) (5 pts) Show that the net intensity measured by the photodiode can be written as 1 = 11 [(1 + r)(1 + q) + 2vr (cos ?? + q cos&J), where ?? and ?2 are associated with wavelengths ?1 and ?2. [Hint: think about how ?2 interferes with d2]

Answer 1

The intensity due to is

The intensity due to lambda_1 is i_1 = I_1(1 + r + 2 squareroot r cos delta_1) (1) The intensity due to lambda_2 is i_2 = I_2(1 + r +2 squareroot r cos delta_2) (2) Note that the phase difference is different for the different wavelengths because the phase difference depends on wavelengths. delta_1 = 2 pi / lambda_1 Delta s and delta_2 = 2 pi / lambda_2 Delta s Now note that I_2 = qI_1 Therefore equation (2) becomes i_2 = qI_1(1 + r + 2 squareroot r cos delta_2) = qI_1 (1 + r) + 2 q squareroot r I_1 cos delta_2 So the net intensity is Hence the net intensity measured by the photodiode is I = I_1[(1 + r)(1 + q) + 2 squareroot r(cos delta_1 + q cos delta_2)]

$i_1=I_1(1+r+2\sqrt{r}\cos \delta_1)$ ....(1)

The intensity due to $\lambda_2$ is

$i_2=I_2(1+r+2\sqrt{r}\cos \delta_2)$ ....(2)

Note that the phase difference is different for the different wavelengths because the phase difference depends on wavelengths.

$\delta_1=\frac{2\pi}{\lambda_1}\Delta s$

and

$\delta_2=\frac{2\pi}{\lambda_2}\Delta s$

Now note that

Therefore equation (2) becomes

$i_2=qI_1(1+r+2\sqrt{r}\cos \delta_2)=qI_1(1+r)+2 q\sqrt{r}I_1\cos \delta_2$

So the net intensity is

$\begin{align} I&=i_1+ i_2=I_1(1+r+2\sqrt{r}\cos \delta_1)+qI_1(1+r)+2 qI_1\sqrt{r}\cos \delta_2 \nonumber \nonumber \\ \Rightarrow I &= I_1(1+r)(1+q)+2\sqrt{r}I_1(\cos\delta_1+q\cos \delta_2) \nonumber \end{align}$

Hence the net intensity measured by the photodiode is

$I = I_1\Big[(1+r)(1+q)+2\sqrt{r}(\cos\delta_1+q\cos \delta_2)\Big]$