The Timpson Lyceum

1. (1 pt •) Two loudspeakers driven by the same amplifier emit identical waves of wavelength $\lambda=480\,\text{nm}$ (treat as coherent light sources for this problem). At a certain point the waves from the two sources have traveled path lengths that differ by $\Delta r = 1.44\,\mu\text{m}$ .

Is the interference at that point constructive or destructive? Justify with the ratio $\Delta r/\lambda$ .

Solution

1.

$\dfrac{\Delta r}{\lambda}=\dfrac{1.44\times10^{-6}}{480\times10^{-9}}=3.0$ , a whole number of wavelengths. Since $\Delta r=3\lambda$ , the two waves arrive in phase, so the interference is constructive (this is the $m=3$ maximum).

2. (2 pts ••) In a Young's double-slit experiment the slits are separated by $d=0.20\,\text{mm}$ and the screen is $L=1.8\,\text{m}$ away. Monochromatic light of wavelength $\lambda=550\,\text{nm}$ is used.

(a) Find the spacing between adjacent bright fringes on the screen.

(b) How far from the central maximum is the tenth-order ( $m=10$ ) bright fringe?

Solution

2.

(a) Fringe spacing $\Delta y=\dfrac{\lambda L}{d}=\dfrac{(550\times10^{-9})(1.8)}{0.20\times10^{-3}}=4.95\times10^{-3}\,\text{m}\approx 5.0\,\text{mm}$ .

(b) The $m$ -th bright fringe is at $y_m=m\dfrac{\lambda L}{d}=m\,\Delta y$ , so $y_{10}=10(4.95\,\text{mm})\approx 50\,\text{mm}=5.0\,\text{cm}$ .

3. (2 pts ••) Coherent light of wavelength $\lambda=633\,\text{nm}$ passes through two slits separated by $d=0.12\,\text{mm}$ .

(a) At what angle $\theta$ (measured from the central axis) does the second-order bright fringe appear?

(b) If the slit separation were instead $d=2.5\,\mu\text{m}$ , how many bright fringes in total would appear on a distant screen? (Count all orders that satisfy $|\sin\theta|\le 1$ .)

Solution

3.

(a) Bright fringes: $d\sin\theta=m\lambda$ . For $m=2$ : $\sin\theta=\dfrac{2\lambda}{d}=\dfrac{2(633\times10^{-9})}{0.12\times10^{-3}}=1.055\times10^{-2}$ , so $\theta=\sin^{-1}(0.01055)\approx 0.60^\circ$ .

(b) The highest order satisfies $\sin\theta\le1$ , i.e. $m_{\max}=\big\lfloor d/\lambda\big\rfloor$ . Here $\dfrac{d}{\lambda}=\dfrac{2.5\times10^{-6}}{633\times10^{-9}}=3.95$ , so $m_{\max}=3$ . Orders $m=0,\pm1,\pm2,\pm3$ appear: total $=2(3)+1=\mathbf{7}$ bright fringes.

4. (2 pts ••) Two identical coherent sources illuminate a screen; at the center of the pattern (zero path difference) the combined intensity is $I_0$ . At a point $P$ their waves arrive with a path difference $\Delta r$ , giving a combined intensity $I=I_0\cos^2(\phi/2)$ , where $\phi$ is the phase difference.

(a) Find the phase difference $\phi$ (in radians and degrees) when $\Delta r = 0.10\,\lambda$ , and evaluate $I/I_0$ there.

(b) For what smallest nonzero path difference (as a fraction of $\lambda$ ) does the intensity fall to one-quarter of its central-maximum value?

Solution

4.

(a) Phase difference $\phi=\dfrac{2\pi}{\lambda}\Delta r=2\pi(0.10)=0.63\,\text{rad}=36^\circ$ . Then $I/I_0=\cos^2(\phi/2)=\cos^2(18^\circ)=0.90$ , so $I\approx0.90\,I_0$ .

(b) Want $\cos^2(\phi/2)=\tfrac14\Rightarrow\cos(\phi/2)=\tfrac12\Rightarrow \phi/2=60^\circ\Rightarrow\phi=120^\circ=\dfrac{2\pi}{3}$ . Since $\phi=2\pi\,\Delta r/\lambda$ , this is $\Delta r=\dfrac{\phi}{2\pi}\lambda=\dfrac{\lambda}{3}$ .

5. (2 pts ••) A soap film ( $n=1.33$ ) of uniform thickness $t=320\,\text{nm}$ is surrounded by air and illuminated at normal incidence with white light.

(a) Explain why light reflected from the front surface undergoes a half-cycle ( $\pi$ ) phase shift while light reflected from the back surface does not.

(b) Which visible wavelength ( $400$ – $700\,\text{nm}$ ) is most strongly reflected? Identify its color.

Solution

5.

(a) Reflection at a boundary where the light goes into a medium of higher refractive index produces a half-cycle ( $\pi$ ) phase shift. At the front surface light goes air ( $n=1$ ) → film ( $n=1.33$ ), higher index, so a $\pi$ shift occurs. At the back surface light goes film ( $n=1.33$ ) → air ( $n=1$ ), lower index, so there is no shift.

(b) With one $\pi$ shift, the extra half-wavelength must be compensated for the two reflected rays to end up in phase (constructive/strong reflection): $2nt=(m+\tfrac12)\lambda$ , i.e. $\lambda=\dfrac{2nt}{m+\frac12}$ . With $2nt=2(1.33)(320\,\text{nm})=851\,\text{nm}$ : $m=0:\ 1702\,\text{nm}$ (IR); $m=1:\ 567\,\text{nm}$ ; $m=2:\ 340\,\text{nm}$ (UV). Only $m=1$ lands in the visible: $\lambda\approx 5.7\times10^{2}\,\text{nm}$ — green.

6. (2 pts ••) A lens is to be coated with a single layer of magnesium fluoride ( $n=1.38$ ) to minimize reflection of $\lambda=550\,\text{nm}$ light at normal incidence. The lens glass has $n=1.52$ .

(a) Both reflections (air→coating and coating→glass) occur at a surface of higher index, so each reflected ray gets a $\pi$ phase shift. Explain whether these two shifts add or cancel, and use that to write the correct condition on $2nt$ for destructive interference (a dark, non-reflecting result) of the reflected light.

(b) Find the minimum (nonzero) coating thickness.

Solution

6.

(a) Because each reflected ray suffers a $\pi$ shift, the two shifts are equal and cancel — they contribute no net relative phase. The only relative phase then comes from the extra path $2nt$ traveled in the film. For the reflected light to interfere destructively (minimum reflection), that path must be a half-integer number of wavelengths:

2nt=\left(m+\tfrac12\right)\lambda,\qquad m=0,1,2,\dots

(b) Minimum nonzero thickness is $m=0$ : $t=\dfrac{\lambda}{4n}=\dfrac{550\,\text{nm}}{4(1.38)}\approx 1.0\times10^{2}\,\text{nm}$ (about $100\,\text{nm}$ ).

7. (3 pts •••) In a Newton's-rings apparatus a plano-convex lens of radius of curvature $R=10.0\,\text{m}$ rests on a flat glass plate, and the gap is illuminated from above with sodium light, $\lambda=589\,\text{nm}$ . Viewed in reflection, a pattern of concentric dark rings appears.

(a) By considering the two reflected rays from the top and bottom of the air gap, show that the dark rings occur where the gap thickness satisfies $2t=m\lambda$ ( $m=0,1,2,\dots$ ), and hence that the radius of the $m$ -th dark ring is $r_m=\sqrt{m\lambda R}$ . (Use $t \approx r^2/2R$ for the thin gap.)

(b) Compute the radius of the fifth dark ring ( $m=5$ ).

Solution

7.

(a) The top reflection (air→glass lens surface, but the ray of interest reflects off the bottom of the air gap: air→flat glass plate, higher index) gets a $\pi$ shift; the reflection off the top of the air gap (glass lens → air, lower index) gets none. So there is a net half-wavelength of relative phase from the reflections alone. The path difference from crossing the gap twice is $2t$ . Combining, the reflected rays interfere destructively (dark) when

2t=m\lambda,\qquad m=0,1,2,\dots

(the extra $\pi$ from reflection already supplies the "half," so the path term must be an integer number of wavelengths). Geometry of a sphere of radius $R$ over a flat plate gives $t\approx \dfrac{r^2}{2R}$ for $r\ll R$ . Substituting:

2\cdot\frac{r^2}{2R}=m\lambda\ \Rightarrow\ r_m=\sqrt{m\lambda R}.

(b) $r_5=\sqrt{5(589\times10^{-9})(10.0)}=\sqrt{2.945\times10^{-5}}\,\text{m}=5.43\times10^{-3}\,\text{m}\approx 5.4\,\text{mm}$ .

Extra Credit (3 pts •••)

Two slits act as identical coherent sources. Each alone produces at a point $P$ on a distant screen a field of amplitude $E_0$ , i.e. $E_1(t)=E_0\cos\omega t$ and $E_2(t)=E_0\cos(\omega t+\phi)$ , where $\phi$ is the phase difference from the path-length difference between the two slits and $P$ .

(a) By adding the two waves — either with the trigonometric sum-to-product identity or by adding phasors (vectors of length $E_0$ separated by angle $\phi$ ) — show that the resultant field is a single cosine of amplitude

E_P = 2E_0\cos\!\left(\frac{\phi}{2}\right).

(b) Intensity is proportional to the square of the field amplitude. Using the fact that a single source alone gives $I_0 \propto E_0^2$ , derive

I = I_0\cos^2\!\left(\frac{\phi}{2}\right),

and identify the value of $I$ at the maxima in terms of $I_0$ . Explain in one sentence why this is not a violation of energy conservation.

(c) Relate $\phi$ to the path difference $\Delta r$ and, for small angles in a double-slit geometry ( $\Delta r = d\sin\theta \approx d\,y/L$ ), show that this intensity formula reproduces bright fringes at $y_m=m\lambda L/d$ .

Solution

Extra Credit

(a) Add the two fields:

E_P=E_1+E_2=E_0\cos\omega t+E_0\cos(\omega t+\phi).

Using $\cos A+\cos B=2\cos\!\frac{A+B}{2}\cos\!\frac{A-B}{2}$ with $A=\omega t$ , $B=\omega t+\phi$ :

E_P=2E_0\cos\!\left(\frac{\phi}{2}\right)\cos\!\left(\omega t+\frac{\phi}{2}\right).

The time-varying factor $\cos(\omega t+\phi/2)$ is a single oscillation at the original frequency; its amplitude is

E_P^{\text{(amp)}}=2E_0\cos\!\left(\frac{\phi}{2}\right).

(Phasor picture: two vectors of length $E_0$ with an angle $\phi$ between them form an isosceles triangle; the resultant bisects the angle and has length $2E_0\cos(\phi/2)$ — the same result.)

(b) Intensity $\propto$ (amplitude) $^2$ . Write $I=C\,\big[2E_0\cos(\phi/2)\big]^2=4CE_0^2\cos^2(\phi/2)$ . A single source alone gives amplitude $E_0$ and intensity $I_0=CE_0^2$ . Therefore

I=4I_0\cos^2\!\left(\frac{\phi}{2}\right).

Writing $I_{\max}\equiv 4I_0$ (the value at $\phi=0$ , where the amplitudes fully add), this is $I=I_{\max}\cos^2(\phi/2)$ , the form quoted in the problems (with $I_0$ there meaning the maximum two-source intensity). At the maxima $I=4I_0$ — four times what one slit gives, not twice.

Energy is still conserved: the extra intensity at the bright fringes is exactly balanced by the zero intensity at the dark fringes. Averaging $\cos^2(\phi/2)$ over the screen gives $\langle I\rangle=\tfrac12(4I_0)=2I_0$ , the sum of the two independent sources' powers — interference merely redistributes the same total energy.

(c) A path difference $\Delta r$ corresponds to phase $\phi=\dfrac{2\pi}{\lambda}\Delta r$ . Maxima occur when $\cos^2(\phi/2)=1$ , i.e. $\phi=2\pi m$ , giving $\Delta r=m\lambda$ . In the double-slit geometry $\Delta r=d\sin\theta\approx d\,y/L$ for small angles, so

\frac{d\,y}{L}=m\lambda\ \Rightarrow\ y_m=\frac{m\lambda L}{d},

the standard bright-fringe positions.

Chapter 35 — Interference

Chapter 35 — Interference

Extra Credit (3 pts •••)

Extra Credit