3604 Multiple Slits

Exercises

21a, 22

36.21 An interference pattern is produced by light of wave length 580 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.530 mm. (a) If the slits are very narrow, what would be the angular positions of the first-order and second-order, two-slit interference maxima?

Solution

dsin⁡θ=mλsin⁡θ=mλdθ1=arcsin⁡(1×5.8×10−70.53×10−3)=0.0627°θ2=arcsin⁡(2×5.8×10−70.53×10−3)=0.125°\begin{aligned} d\sin \th &= m \lambda\\ \sin \th &= \frac{m \lambda}{d}\\ \th_1 &= \arcsin(\frac{1 \times 5.8 \times 10^{-7}}{0.53 \times 10^-3}) = 0.0627 \degree\\ \th_2 &= \arcsin(\frac{2 \times 5.8 \times 10^{-7}}{0.53 \times 10^-3}) = 0.125 \degree\\ \end{aligned}

36.22 Laser light of wavelength 500.0 nm illuminates two identical slits, producing an interference pattern on a screen 90.0 cm from the slits. The bright bands are 1.00 cm apart, and the third bright bands on either side of the central maximum are missing in the pattern. Find the width and the separation of the two slits.

Solution

yR=tan⁡θ≈sin⁡θ=mλd⇒y=RmλdΔy=ym+1−ym=λRdd=λRΔyλ=500 nm,R=90 cm,Δy=1 cm⇒d=4.5×10−5 m\begin{aligned} \frac{y}{R} &= \tan \th \approx \sin \th = \frac{m \lambda}{d}\\ \To y& = R\frac{m \lambda}{d}\\ \Delta y &= y_{m+1} - y_m = \frac{\lambda R}{d}\\ d &= \frac{\lambda R}{\Delta y}\\ \lambda &= 500 \text{ nm}, R = 90 \text{ cm}, \Delta y = 1 \text{ cm}\\ \To d &= 4.5 \times 10^{-5} \text{ m} \end{aligned}

Since the third bright band is missing, then d=3a⇒a=1.5×10−5 md = 3a \To a = 1.5 \times 10^{-5} \text{ m} .
If the third interference bright fringe is missing from the pattern, this means it has the same position as the first diffraction minimum, therefore,

sin⁡θint=mλdsin⁡θdiff=m′λa=λaθint=θdiff3λd=λa⇒d=3a\begin{aligned} \sin\th_{int} &= \frac{m \lambda}{d}\\ \sin\th_{diff} &= \frac{m' \lambda}{a} = \frac{\lambda}{a}\\ \th_{int} &= \th_{diff}\\ \frac{3 \lambda}{d} &= \frac{\lambda}{a}\\ \To d& = 3a \end{aligned}