Problem 12.2.11^∗

Solution

Data:

Parallel beam of monochromatic light of wavelength $\lambda$

Observation point A on the axis.

The plate has rings that coincide with the Fresnel zones.

It is considered that the plate covers the first m = 9 Fresnel zones.

Reference amplitude (without plate):
With the entire wavefront open, the amplitude at A is half of the first zone:

$A_{\text{free}} \approx \frac{A_1}{2}$

and the corresponding intensity is

$I_0 \propto \left(\frac{A_1}{2}\right)^2 = \frac{A_1^2}{4}$

Case (a): Opaque plate on even zones

The plate has the even zones blackened out (blocked): 2, 4, 6, 8.
The odd zones remain open: 1, 3, 5, 7, 9 (five zones).

Since the amplitudes of the zones are

approximately equal
$(A_1 \approx A_3 \approx A_5 \approx A_7 \approx A_9)$, the total amplitude at A is:

$A_a = 5 A_1$

Intensity:

$I_a \propto (5 A_1)^2 = 25 A_1^2$

Compared with the intensity without the plate:

$\frac{I_a}{I_0} = \frac{25 A_1^2}{A_1^2/4} = 25 \times 4 = 100$

$\boxed{\text{The intensity increases by a factor of 100.}}$

Case (b): Phase plate (phase shift \pi) on even zones

The rings that were previously opaque are replaced by dielectric layers that add a phase shift of \pi to the wave passing through.

· The odd zones (1, 3, 5, 7, 9) pass with no phase change.
· The even zones (2, 4, 6, 8) would normally arrive with a phase shift of $\pi$due to the path difference, but the layer adds another \pi, so they end up in phase with the odd ones.

Now all zones (9 in total) contribute constructively:

$A_b = 9 A_1$

Intensity:

$I_b \propto (9 A_1)^2 = 81 A_1^2$

Compared with the intensity without the plate:

$\frac{I_b}{I_0} = \frac{81 A_1^2}{A_1^2/4} = 81 \times 4 = 324$

$\boxed{\text{The intensity increases by a factor of 324.}}$