Intensity in Double-Slit Interference — Question 3

Question 3

A neutral-density filter attenuates the field by the square root of its transmission

Two initially equal coherent beams each have intensity $I_0$. An ideal neutral-density filter with intensity transmission $T=0.25$ is placed over one slit; it adds no phase shift. The polarizations remain parallel.

1. Calculate the extrema and visibility after inserting the filter.

2. Find the total intensity at quadrature and compare the phase-averaged intensity before and after filtering.

Question 3 - Solution

A neutral-density filter attenuates the field by the square root of its transmission

(a) Method and calculation

$$E_2^\prime=\sqrt T E_0=E_0/2,\quad I/I_0=1+T+2\sqrt T\cos\delta.$$$$\boxed{I_{\max}=2.25I_0,\quad I_{\min}=0.25I_0,\quad V=\frac{2\sqrt T}{1+T}=0.80}.$$

(b) Interpretation and further calculation

$$\delta=\pi/2:\quad\boxed{I=1.25I_0}.$$$$\boxed{\langle I\rangle_{\mathrm{before}}=2I_0,\quad\langle I\rangle_{\mathrm{after}}=(1+T)I_0=1.25I_0}.$$

Physical check

Using T rather than its square root on the field amplitude would overestimate the loss of interference modulation.