Young's Interference Experiment — Question 1

Question 1

Exact screen geometry reveals the error in uniform fringe spacing

Two narrow slits separated by $d=10.0\,\mu\mathrm m$ are illuminated normally with $\lambda=500\,\mathrm{nm}$. A screen is $L=1.00\,\mathrm m$ away. Use the far-field path difference $d\sin\theta$, but retain the exact screen relation $y=L\tan\theta$.

1. Find the position of the fourth bright fringe above the axis.

2. Compare with $y_4\simeq4\lambda L/d$ and calculate the fractional underestimate relative to the exact result.

Question 1 - Solution

Exact screen geometry reveals the error in uniform fringe spacing

(a) Method and calculation

$$d\sin\theta_m=m\lambda\Longrightarrow y_m=\frac{Lm\lambda/d}{\sqrt{1-(m\lambda/d)^2}}.$$$$\boxed{y_4=\frac{0.200}{\sqrt{0.96}}\,\mathrm m=0.204124\,\mathrm m}.$$

(b) Interpretation and further calculation

$$y_{4,\mathrm{parax}}=0.200000\,\mathrm m,\qquad y_4-y_{4,\mathrm{parax}}=4.124\,\mathrm{mm}.$$$$\boxed{\epsilon=1-\frac{y_{4,\mathrm{parax}}}{y_4}=1-\sqrt{0.96}=2.0204\%}.$$

Physical check

Equal increments in sine of angle do not produce exactly equal increments of screen height at large angles.