Michelson's Interferometer — Question 5

Question 5

Equal-inclination rings are uniformly spaced in squared radius

The equivalent reflecting planes in an aligned Michelson are separated by $d=1.00\,\mathrm{mm}$. Ignore a fixed phase offset, so bright directions obey $2d\cos\theta=m\lambda$. Let $\lambda=500\,\mathrm{nm}$ and observe through a lens of focal length $f=0.200\,\mathrm m$. The axial order is the integer $m_0=2d/\lambda$.

1. Using $\cos\theta\simeq1-\theta^2/2$ and $r\simeq f\theta$, derive the radius of the bright ring p orders below the axial order.

2. Find the first and third ring radii and the change in their radii if d doubles while f and wavelength are fixed.

Question 5 - Solution

Equal-inclination rings are uniformly spaced in squared radius

(a) Method and calculation

$$2d-d\theta_p^2=(m_0-p)\lambda=2d-p\lambda\Longrightarrow\theta_p^2=p\lambda/d.$$$$\boxed{r_p^2=f^2p\lambda/d,\quad r_{p+1}^2-r_p^2=f^2\lambda/d}.$$

(b) Interpretation and further calculation

$$\boxed{r_1=4.47214\,\mathrm{mm},\quad r_3=7.74597\,\mathrm{mm}}.$$$$d\to2d\Longrightarrow\boxed{r_p\to r_p/\sqrt2};\quad m_0:4000\to8000.$$

Physical check

The ring radius is proportional to the square root of order difference, so adjacent rings crowd together at larger radii.