Homework Solutions

286
Just count waves

\frac{d sin θ}{\frac{λ}{n}} - \frac{d sin 5^{o}}{λ} = m = 1, sin θ = \frac{λ + d sin 5^{o}}{n d} = 0.066, θ = 3.7 8^{o}, y = L tan θ = 0.33 m

287

a sin θ = a \frac{4 \times 1 0^{- 4} m}{4 m} = 1 \cdot 6.0 \times 1 0^{- 7} m, a = 6.0 \times 1 0^{- 4} m, d = 4 a = 2.4 \times 1 0^{- 3} m

288

d sin θ_{3} \approx d tan θ_{3} = d \frac{y_{3}}{L} = 3 λ, d sin θ_{5} \approx d tan θ_{5} = d \frac{y_{5}}{L} = 5 λ

y_{5} - y_{3} = \frac{2 λ L}{d} = 2 \cdot 5 \times 1 0^{- 7} m \cdot \frac{2.0 m}{1 \times 1 0^{- 4} m}

289
Just look at the ﬁgure above for which $d = 4 a$ , there are $7 = 2 \cdot 4 - 1$ fringes between $m = - 1$ and $m = 1$ diffraction zeros, and $3 = 4 - 1$ between $m = 2$ and $m = 1$ , so if $d = 5 a$ there will be $2 \cdot 5 - 1 = 9$ between $m = - 1$ and $m = 1$ diffraction zeros, and $4 = 5 - 1$ between $m = 2$ and $m = 1$ .

290
Since $d sin θ \approx d tan θ = d \frac{y}{L} = m λ$ , we see that cutting $L$ in half will cut $y$ in half if all else stays the same. Since we are dealing with waves emanating from slits (effectively line sources) we must have $E \propto \frac{1}{\sqrt{r}}$ so that $I \propto \frac{1}{r}$ , since $I = \frac{℘}{2 π r ℓ}$ , therefore moving the screen to half as far away will double the light intensity arriving at any point on the screen. If these were point-sources of light, you would quadruple the intensity.

291

(\frac{sin 3 δ}{sin δ})^{2} = (\frac{sin 2 δ cos δ + cos 2 δ sin δ}{sin δ})^{2} = (\frac{2 sin δ {cos}^{2} δ + cos 2 δ sin δ}{sin δ})^{2} = (2 {cos}^{2} δ + cos 2 δ)^{2}

= (1 + 2 cos 2 δ)^{2} = 1 + 4 cos 2 δ + 4 {cos}^{2} 2 δ, δ = \frac{π d sin θ}{λ}

292

{cos}^{2} \frac{π d sin θ}{λ} (\frac{sin \frac{π a sin θ}{λ}}{\frac{π a sin θ}{λ}})^{2} \to {cos}^{2} \frac{π a sin θ}{λ} (\frac{sin \frac{π a sin θ}{λ}}{\frac{π a sin θ}{λ}})^{2}

= (\frac{sin \frac{π a sin θ}{λ} cos \frac{π a sin θ}{λ}}{\frac{π a sin θ}{λ}})^{2} = (2 \frac{sin \frac{π a sin θ}{λ} cos \frac{π a sin θ}{λ}}{\frac{2 π a sin θ}{λ}})^{2} = (\frac{sin \frac{π 2 a sin θ}{λ}}{\frac{π 2 a sin θ}{λ}})^{2}

293
It is as if there were a second virtual slit (the image of the real slit in the mirror) a distance below the mirror, so that maxima should be at $2 d sin θ = m λ$ and $I = I_{0} cos (\frac{π 2 d sin θ}{λ})$ , however the extra ﬂip of electric ﬁeld due to the reﬂection turns the condition for a maximum into that for a dark fringe, making $I = I_{0} {sin}^{2} (\frac{π 2 d sin θ}{λ})$ with maxima at $2 d sin θ = (m + \frac{1}{2}) λ$ .
We can get this from adding the waves

E = E_{0} p sin (k r_{-} - ω t) + (- E_{0} p) sin (k r_{+} - ω t) = 2 E_{0} p cos (k (\frac{r_{-} + r_{+}}{2}) - ω t) sin (k (\frac{r_{-} - r_{+}}{2}))

square, time average;

Ī = \frac{4 E_{0}^{2}}{μ_{0} c} \frac{1}{2} {sin}^{2} (k (\frac{r_{-} - r_{+}}{2})) = I_{0} {sin}^{2} (\frac{π 2 d sin θ}{λ})

294
The fringes visibly seperated at the $m = 3$ (with $m = 0$ in the middle), so resolution occurs in third order. There are $8$ tiny secondary fringes between adjacent superfringes, so $N_{s l i t} = 8 + 2 = 10$ . Look at the third order seperated fringes;

d sin θ_{1} = 3 λ_{1} = d \frac{x_{1}}{L}, d sin θ_{2} = 3 λ_{2} = d \frac{x_{2}}{L}

\frac{λ_{1}}{λ_{2}} = \frac{x_{1}}{x_{2}} = \frac{0.038}{0.04}

so $λ_{2} = 650.75 n m$ or $721.05 n m$ since you do not know which $λ$ produced which set of fringes.

295
There is one ﬂip at the bottom of the air layer;

2 n d = m λ, 2 \cdot 1 \cdot 1.0 \times 1 0^{- 4} m = m \cdot 5.0 \times 1 0^{- 7} m, m = 400

If you include the dark disk at the the center ( $m = 0$ ) there are $401$ rings, but $400$ are true circles. Either answer is acceptable.
The ring radii are in proportion to the ﬁlm thickness here since the air layer is a triangle in cross section, at ring $m = 10$ ;

2 \cdot 1 \cdot d_{10} = 10 \cdot 5.0 \times 1 0^{- 7} m, d_{10} = 2.5 \times 1 0^{- 6} m

\frac{r_{10}}{r_{400}} = \frac{d_{10}}{d_{400}}, r_{10} = 2.0 \times 1 0^{- 2} m \cdot \frac{2.5 \times 1 0^{- 6} m}{1.0 \times 1 0^{- 4} m} = 5.0 \times 1 0^{- 4} m

296
We saw that $r_{m} = \sqrt{m λ R}$ , so $r_{m + 1} = \sqrt{(m + 1) λ R}$

\frac{d r_{m}}{d m} = \frac{1}{2} \sqrt{\frac{R λ}{m}} \approx \frac{r_{m + 1} - r_{m}}{(m + 1) - m}

297

r_{n}^{2} = n λ R = {(0.164 c m)}^{2}, r_{n + 20}^{2} = (n + 20) R λ = {(0.368 c m)}^{2}, {(0.368 c m)}^{2} - {(0.164 c m)}^{2} = 20 R 5.46 \times 1 0^{- 7} m

R = \frac{{(0.00368 m)}^{2} - {(0.00164 m)}^{2}}{20 \cdot 5.46 \times 1 0^{- 7} m}

298
The surfaces perform one inversion, so

2 \cdot (1.33) d = m \cdot 4.5 \times 1 0^{- 7}, 2 \cdot (1.33) d = (m^{'} + \frac{1}{2}) \cdot 6.0 \times 1 0^{- 7}

One solution is $m = 2$ , $m^{'} = 1$ . For these numbers $d = \frac{900 n m}{2.66} = 338 n m$ , there will be minima for $λ = \frac{900 n m}{1}$ and $λ = \frac{900 n m}{3}$ and these do not lie between $450$ and $600 n m$ so we are OK. Another solution is $m = 6$ and $m^{'} = 4$ , but this makes $d = 1015 n m$ , and there will be minima for

λ = \frac{2700 n m}{5}, \frac{2700 n m}{4}, \frac{2700 n m}{3}, \frac{2700 n m}{2}, \frac{2700 n m}{1}

and one of these is between $450$ and $600 n m$ , and so we toss out this and higher solutions and stick with the ﬁrst solution.