Intensity in Single-Slit Diffraction, Qualitatively — Question 5

Question 5

Holding transmitted power fixed changes how the central peak scales

A uniform slit of width a has field amplitude $A_0$. In a one-dimensional paraxial model, transmitted power is proportional to $A_0^2a$, central far-field intensity to $A_0^2a^2$, and angular width to $1/a$. The width doubles while the incident field is adjusted to keep transmitted power fixed.

1. Find the required field-amplitude change and the central-intensity ratio.

2. Find the angular-width ratio and verify consistency with the integrated-power constraint.

Question 5 - Solution

Holding transmitted power fixed changes how the central peak scales

(a) Method and calculation

$$A_0^{\prime2}(2a)=A_0^2a\Longrightarrow\boxed{A_0^\prime=A_0/\sqrt2}.$$$$\boxed{I_c^\prime/I_c=(1/2)(2^2)=2}.$$

(b) Interpretation and further calculation

$$\boxed{\Delta\theta^\prime/\Delta\theta=1/2}.$$$$\boxed{(I_c^\prime/I_c)(\Delta\theta^\prime/\Delta\theta)=2(1/2)=1}.$$

Physical check

This fixed-power comparison differs from keeping the incident irradiance fixed, which would increase transmitted power as the slit widens.