GR 8677927796770177 | # Login | Register

All Solutions of Type: Optics
 0 Click here to jump to the problem! GR8677 #13 Optics$\Rightarrow$}Interference Fact: The human eye can only see things in motion up to about 25 Hz. (One can approximate this knowing that the human eye blinks once every three seconds on average.) Now, the problem mentions that the relative phase is varied, at a constant frequency of 500 Hz, which is much greater than the maximum frequency of the human eye. Interference is produced as long as the sources are coherent, and the sources are coherent as long as there's a constant relation between relative phase in time. (A) The frequency of the phase change has nothing to do with the color of light. (B) Interference pattern is different for $\pi$ and $2\pi$ phase changes... (C) Interference can exist for other phase differences. (D) One can have interference even with polychromatic sources. (E) The interference pattern shifts position (since the source remains coherent from the constant relation with relative phase to time) at a rate too fast for the human eye, as explained above. Click here to jump to the problem!
 1 Click here to jump to the problem! GR8677 #58 Optics$\Rightarrow$}Holograms A hologram is produced from interference of light. Interference of two-beams (one going directly to the film, the other bouncing off the object) produced by a beam-splitter allows the film to record both intensity and relative phase of the light at each point. Intensity (to wit: $I\propto E_0^2 \langle \sin^2 \omega t \rangle=E_0^2/2$) does not depend on angular frequency, but only on phase and amplitude. Thus, choice (B) is right. Click here to jump to the problem!
 2 Click here to jump to the problem! GR8677 #73 Optics$\Rightarrow$}Thin film In order for the thin film layer to be non-reflecting, it must cancel the reflected wavelengths---as in a destructive interference. The change in wavelengths is $\Delta \lambda = \lambda - \lambda/2 = \lambda /2$, since the wave changes phase by $\pi$ at the interface between air and the coating, and changes phase again at the second interface between coating and glass. (Assume that $n_{air} < n_{coating} < n_{glass}$.) Destructive interference is thus given by a half-integer wavelength change $m \lambda /2$, where the smallest change is $\lambda/2$. The change in wavelength occurs over twice the thickness $t$ of the coating, thus $2t = \Delta \lambda = \lambda/2. $ This implies that $t =\lambda/4$, as in choice (A). Click here to jump to the problem!
 3 Click here to jump to the problem! GR8677 #74 Optics$\Rightarrow$}Polarizers One might remember the result from optics that the maximum fraction incident between three polarizers with the first and third orthogonal to each other is $1/8$. Or, if not, one can derive it rather quickly: Suppose the incident intensity of the light (before going through any polarizers) is $I$. Light going through the first polarizer has the intensity $I_1=I/2$. Light going through the second polarizer has the intensity $I_2=I_1 \cos^2 \phi =I/2 \cos^2 \phi$. ($\phi$ is the angle between the polarizer and the light.) Light going through the third polarizer has the intensity $I_3=I_2 \cos^2 \theta = I/2 \cos^2 \phi \cos^2 \theta$. ($\theta$ is the angle between the polarizer and the light.) In order for the intensity $I_3$ to be max, one can take the derivative with respect to either $\theta$ or $\phi$. Knowing a priori that the first and third polarizers are orthogonal (at 90 degree angles) to each other, one can rewrite either $\phi$ or $\theta$ in terms of the other. So, $\phi = \pi/2 - \theta$, and thus, $I_3 = I/2 \cos^2 (\pi/2 - \theta) \cos^2 (\theta)= I/2 \sin^2\theta \cos^2\theta = I/4 \sin(2\theta).$ The the derivative to find the maximum, $\frac{d I_3}{d \theta}=I/2\cos(2\theta)=0$. One finds that $2\theta = \pi/2 \Rightarrow \theta = \pi/4$. This implies that Plug in $\theta$ to get $I_3=1/8$. Click here to jump to the problem!
 4 Click here to jump to the problem! GR8677 #91 Optics$\Rightarrow$}Bragg Reflection Bragg diffraction is basically the wavelength change of a wave impinging two adjacent layers of the crystal. More specifically, it relates the wavelength difference of the incident wave on the top layer to that of the reflected wave on the lower layer. The distance between layers (lattice planes", in Solid-State-Speak) is $d$, and the angle of incidence is $\theta$. The change in wavelength is thus a simple geometry problem, the result being the celebrated Bragg Reflection Relation, $2d \sin \theta = n \lambda, $ where $n$ corresponds to the order of the reflection. The problem supplies the following numbers $\begin{eqnarray} n=1\\ d=3E-10\\ \theta = 30^\circ\\ m_e\approx 10E-31, \end{eqnarray}$ and if one recalls the de Broglie relation, $p = h/\lambda$, one has, in general, $n\lambda = nh/{m v} = 2d \sin\theta$. Plugging in some numbers, one finds that $\frac{6E-34}{10E-31v}=3E-10$, since $\sin 30^\circ = 1/2$, and where the approximation scheme to battle the no-calculators-allow requirement is shown. Thus, one has $v=0.25E7$, which is very close to choice (D). The problem . Click here to jump to the problem!
 5 Click here to jump to the problem! GR8677 #100 Optics$\Rightarrow$}Diffraction The key equation involved is the (Fraunhoffer) diffraction equation, $\sin \theta = \lambda/d$, where the obvious quantities are used. For small angle, the equation simplifies to $\theta \approx \lambda/d$. The blur would be due to both the size of the hole and the diffraction. The arclength of the diffraction is approximated by $d_2=D(2\theta) = 2D\lambda/d$, where the factor of 2 comes from the fact that the arclength angle is symmetrical about the diffraction axis, thus twice the diffraction angle. Define a blur equation $D'=d+d_2=d+2D\lambda/d$. Take the first derivative, with respect to the pinhole diameter, set it to 0, to get $dD'/dd = 1-2D\lambda/d^2=0\Rightarrow d=\sqrt{2D\lambda}\approx \sqrt{D\lambda}$, as in choice (A). \par This solution is due to: \begin{verbatim} http://www.exo.net/~pauld/summer_institute /summer_day3eye_and_brain/pinhole_optimum_size.html \end{verbatim} Click here to jump to the problem!
 6 Click here to jump to the problem! GR8677 #54 Optics$\Rightarrow$}Field Trajectory The problem gives equi-amplitude, thus the field becomes $\vec{E}=Ee^{i(kz-\omega t)}\hat{x} + E e^{i(kz-\omega t + \pi)}\hat{y}$. Taking the real part, (applying Euler's Theorem, to wit: $e^{i\theta}=\cos\theta + i\sin\theta$) one has $\vec{E}=E\cos(kz-\omega t)\hat{x} + E \cos(kz-\omega t + \pi)\hat{y}$. Apply the trig identity $\cos(\alpha \pm \beta)=\cos \alpha \cos \beta \mp \sin\alpha\sin\beta$ to make the field argument equi-phase, $\vec{E}=E\cos(kz-\omega t)\hat{x} - E \cos(kz-\omega t)\hat{y}$. Looking down from the z-axis, one has $z=0\Rightarrow$ $\vec{E}=E\cos(\omega t)\hat{x} - E \cos(\omega t)\hat{y}$. Make a table of a few values of t and E, $\begin{eqnarray} \omega t & \Rightarrow & (\cos(\omega t),-\cos(\omega t))\\ 0 & \Rightarrow & (1,-1)\\ \pi/6 &\Rightarrow& \frac{1}{2}(\sqrt{3},-\sqrt{3})\\ \pi/4 &\Rightarrow& \frac{1}{2}(\sqrt{2},-\sqrt{2}), \end{eqnarray}$ and one deduces that the points plot out a diagonal line at $135^\circ$ to the x-axis, as in choice (B). Click here to jump to the problem!
 7 Click here to jump to the problem! GR8677 #55 Optics$\Rightarrow$}Polarizations After the wave has been de-coupled into separate directions, the intensity adds separately. That is, the intensity of the wave split by the x-polarizer is $I_1=|E_1|^2$, while that of the wave split by the y-polarizer is $I_2=|E_2|^2$. Add the two intensities to get choice (A). Click here to jump to the problem!
 8 Click here to jump to the problem! GR8677 #56 Optics$\Rightarrow$}Total Internal Reflections Total internal reflection is when one has a beam of light having all of the incident wave reflected. Going through a bit of formalism in electromagnetism one can derive Snell's Law for Total Internal Reflection, $n_{inside} \sin\theta = n_{outside},$ where $n_{inside}=1.33$, and one assumes that the surface has $n_{outside}=1$ for air. One must solve the equation $\theta = \sin^{-1}(1/1.33)$. One can immediately throw out choices (A) and (E). From the unit circle, one recalls that $\sin(30^{\deg})=1/2$ and $\sin(60^{\deg})=1.7/2=0.85$. Since $1/1.33 \approx 0.7$, one deduces that the angle must be choice (C). Click here to jump to the problem!
 9 Click here to jump to the problem! GR8677 #57 Optics$\Rightarrow$}Diffractions The single slit diffraction formula is $d \sin \theta = \lambda m$, where one has integer m for maxima and half-integers for minima. (Opposite to single-slit interference.) Given $m=1$, $\theta = 4E-3 rad$, $\lambda = 400E-9 m$, and making the approximation $\sin \theta \approx \theta$ for small angles, one has the following equation for d, $d\approx \frac{\lambda m}{\theta} = \frac{4E-7}{4E-3}=1E-4$, as in choice (C). Click here to jump to the problem!
 10 Click here to jump to the problem! GR8677 #58 Optics$\Rightarrow$}Lensmaker Equations Although this problem mentions lasers, no knowledge of quantum mechanics or even lasers is required. Instead, the problem can be solved as a simple geometric optics problem using the lensmaker's equation, $1/d_i+1/d_o=1/f$, relating the distances of the object, image, and the focus. Since there are two convex lenses, one can treat the set-up as a telescope. The lens closest to the laser is the objective (with focus $f_o$) and the one closest to the bigger-radius well-collimated beam is the eyepiece $f_e$. The laser-light comes in from $d_{o1}=\infty$, and thus one has $d_{i1}=f_o$, i.e., the image forms at the focal point. Using the telescope equation, one has $M = f_o/f_e = 10 \Rightarrow f_e=f_o/10=15cm$, since one wants a final magnification of 10 (to wit: input beam is 1mm, output beam is 10mm). This narrows down the choices to just (D) and (E). Since the distance between lenses for a telescope (with incoming light coming from infinity) is given by $d_{i1}+d_{o2}=f_o+f_e=16.5$, which relates the distance of the image from the first lens to the distance of the object from the second lens, one arrives at choice (E). (Correction due to user tachyon.) Click here to jump to the problem!
 11 Click here to jump to the problem! GR8677 #82 Optics$\Rightarrow$}Thin films For a thin film of thickness t, one can easily find the condition for interference phenomenon. Since the light has to travel approximately $2t$ to get back to the original incidence interface, one has $2t = m\lambda$. However, since the light changes phase at the interface between air and glass (since glass has a higher index of refraction than air), the condition for constructive interference becomes $2t=m\lambda/2$, where $m \in Odd$. One can create a table to determine the values of $t=m\lambda/4$. $\begin{eqnarray} m & \Rightarrow & t\\ 1 & \Rightarrow & \lambda/4 = 122nm\\ 3 & \Rightarrow & 3\lambda/4 = 366nm\\ 5 & \Rightarrow & 5\lambda/4 = 610nm, \end{eqnarray}$ and so forth... One thus finds that choice (E) is correct. Click here to jump to the problem!
 12 Click here to jump to the problem! GR8677 #11 Optics$\Rightarrow$}Lensmaker Formula The lensmaker's formula is $1/d_o+1/d_i=1/f$. For an image on the opposite side of the light, the image distance is taken as positive. The distance between the object and the first lens is $d_{1o}=40$. $f_1=20$. The lensmaker's formula gives $1/d_{1i}=1/f_1-1/d_{1o}=1/40$. Thus, the image is 40 cm behind the first lens. The first image forms the object for the second lens. The distance of the first image to the second lens is $40-30=10cm$. Since this image is behind the lens (on the other side of the incident geometric light), the convention in the lensmaker's formula takes this as a negative distance. One has, $1/d_{i2}=1/f_2 - 1/d_{o2}=1/10+1/10=2/10=1/5$. Thus, the second image is 5 cm to the right of the second lens. Click here to jump to the problem!
 13 Click here to jump to the problem! GR8677 #12 Optics$\Rightarrow$}Mirror Formula The mirror formula is identical in form to the lensmaker's formula in Problem 11. The sign convention varies for the image distance. If an image distance is on the inside of the mirror, then it is taken as negative. Thus, $1/f=1/d_o+1/d_i\Rightarrow 1/d_i=1/f-1/d_o$. From the diagram, one deduces that $d_o. Thus, one finds that $d_i$ has to be negative. The virtual image has to be inside the mirror. Only choice (V) shows this. Click here to jump to the problem!
 14 Click here to jump to the problem! GR8677 #13 Optics$\Rightarrow$}Aperture Formula The circular aperture formula (a.k.a. Rayleigh Criterion) is given by $\theta = 1.22 \lambda / D$. Plug in the given quantities to get that. (It's a nice formula to memorize, as it's used as common sense in a variety of engineering fields, as well as in remote-sensing, such as satellite-communications.) Click here to jump to the problem!
 15 Click here to jump to the problem! GR8677 #51 Optics$\Rightarrow$}Polarizers When one has three polarizers with the first oriented at a 90 degree angle to the last, the maximum light transmitted is $I_0/8$. In this case, the intensity of light transmitted through the first filter is $I_0/2$, where $I_0$ is the incident light. (Half the light has been canceled by the polarization.) The intensity of the light transmitted through the second filter is $I_2 = |I_1 \cos 45^\circ|^2 = I_0/4$. The intensity of the light transmitted through the third filter is $I_3 = |I_2 \cos 45^\circ|^2 = I_0/8$. This is choice (B). Click here to jump to the problem!
 16 Click here to jump to the problem! GR8677 #69 Optics$\Rightarrow$}Thin Films The incident wavelength changes phase at the air-oil boundary and at the oil-glass boundary. Thus, integer wavelengths produce constructive interference. The incident wavelength travels through $2t$. Thus, $2t = \lambda \Rightarrow \lambda \approx 240 nm$. The closest thickness is 200 nm, so choose choice (B). Additionally, if one wants a better approximation, one can use the equation $v=c/n=\lambda f\Rightarrow \lambda_{'}=\lambda_0/n$ to determine the wavelength of the beam in oil. Thus, one arrives at the exact answer 200nm. (Addition due to user E123.) Click here to jump to the problem!
 17 Click here to jump to the problem! GR8677 #70 Optics$\Rightarrow$}Interference The single-slit interference equation for bright fringes is given by $m\lambda \propto d \sin \theta \approx d \theta$ (for small angles), where d is the width of the slit and m is an integer. Since $c=\lambda \nu$, one can relate the above equation to frequency to get $m c/f \propto d \theta$. Increasing frequency would decrease the angle. Thus, the fringes would get closer together. Increasing the frequency by a factor of 2 would decrease the separation by 2, as in choice (B). Click here to jump to the problem!
 18 Click here to jump to the problem! GR8677 #97 Optics$\Rightarrow$}Refraction From Snell's Law, one obtains $\sin\theta = n(\lambda) \sin\theta^{'}$, since the index of refraction of air is about 1. Now, differentiate both sides with respect to $\lambda$. $\begin{eqnarray} \frac{d}{d\lambda}\sin\theta &=& \frac{d}{d\lambda}(n(\lambda) \sin\theta^{'})\\ 0&=& \frac{dn(\lambda)}{d\lambda} \sin\theta^{'}+(n(\lambda) \cos\theta^{'}) \frac{d\theta^{'}}{d\lambda}\\ \delta \theta^{'} &=& |\tan\theta^{'}/n \frac{dn(\lambda)}{d\lambda} \delta \lambda|, \end{eqnarray}$ which gives choice (E) to be the angular spread. This solution is due to ShyamSunder Regunathan. Click here to jump to the problem!
 19 Click here to jump to the problem! GR8677 #100 Optics$\Rightarrow$}Interferometer This is a nice problem. An interferometer, like its name suggests, has to do with interference. Namely, it splits light-beams through the beam-splitters (BS) half and half, then recombines them at the end---the end result shows interference which varies depending on whether the moveable mirror is placed a quarter of a wavelength away (so that the path difference, which is twice that, is half a wavelength). A fringeshift occurs whenever $d = \lambda$. Thus $m = 2d/\lambda$, where m is the number of fringes and $\lambda$ is the wavelength. Using the information for the red beam, one can find $d=m\lambda/2=85865 \times 632.82/2$. Applying the same equation for the green beam, one finds that $\lambda = 2d/m = 85865 \times 632.82/100000 \approx 540nm$, as in choice (B). Click here to jump to the problem!
 20 Click here to jump to the problem! GR8677 #20 Optics$\Rightarrow$}Missing Fringes Missing fringes in a double-slit interference experiment results when diffraction minima cancel interference maxima. From a bit of phasor analysis, one can derive the diffraction factor $\beta/2=\pi w/\lambda \sin\theta$ and the interference factor $\delta/2=\pi d/\lambda \sin\theta$, where w is the width of the slits and d is the separation (taken from slit centers). The angles belong in the intensity equation given by $I \propto \sin(\beta/2)^2 \cos(\delta/2)^2$. Thus, the condition for a double-slit diffraction minimum is given by $\delta/2=m_d\pi=\pi w/\lambda\sin\theta \Rightarrow m_d\lambda = w \sin\theta$. Also, the condition for interference maximum is given by $\beta/2=m_i\pi=\pi d/\lambda \sin\theta \Rightarrow m_i\lambda = d\sin\theta$. Now, one needs to find the choice that allows for an integer $m_d$. This immediately eliminates choices (A) and (B). But, this leaves choices (C), (D), and (E). Among the remaining choices, there is only one choice that allows for slits that are smaller than the separation. This is choice (D). Take it. Click here to jump to the problem!
 21 Click here to jump to the problem! GR8677 #21 Optics$\Rightarrow$}Thin Film Elimination time. I. Can't be this, since one knows from basic thin-film theory that choice IV is right. (None of the letter choices allow for both choices I and IV.) II. Thin film theory has $2t=\lambda/2$ for constructive interference and $2t=\lambda$ for destructive interference. Thus, the thickness of the film is smaller than that of the light. (Search on the homepage of this site for more on thin film theory---it is explained in the context of other problems.) III. This phase change allows for the half-integer constructive interference. IV. Phase change only occurs when light travels from a medium with lower index of refraction to a medium with higher index of refraction. Since at the back surface, the light would be going from higher to lower index of refraction, there is no phase change. Thus, choice (E). Click here to jump to the problem!
 22 Click here to jump to the problem! GR8677 #22 Optics$\Rightarrow$}Telescope The magnification for a telescope is related to the focal length for the eyepiece and objective by $M=f_o/f_e$. (Note that it is the eye-piece that magnifies it. The objective merely sends an image that's within view of the eye-piece. However, magnification is inversely related to focal length.) The problem gives angular magnification to be $M=10=f_o/f_e \Rightarrow f_e=f_o/10=.1m$. The distance between the objective and eyepiece is the sum of the focal lengths (since the light comes from infinity). $d=f_o+f_e=1.1$m as in choice (D). Click here to jump to the problem!
 23 Click here to jump to the problem! GR8677 #35 Optics$\Rightarrow$}Diffraction Grating Diffraction gratings have the same formula as 2-slit interference, except each slit is (obviously) much smaller. The condition for maximum is given by $d\sin\theta = m\lambda$, relating the width of the slit to the wavelength and angle and order m. The width of each slit is given by the grating $d=(2000 lines/cm \times 100cm/m)^{-1}=0.5E-5 m$. Thus, plugging in the wavelength one has $\sin\theta = \lambda/d = 5200E-10/0.5E-5 \approx 10000E-5 = 1E-1$. Now, the approximations to get rid of the trig function. Since $\theta << 1$, one can approximate $\sin\theta \approx \theta$, where the angle is in radians. Now, convert the angle from radians to degrees. $1E-1 \times 180^{\circ}/\pi = 18/\pi \approx 18/3 = 6^{\circ}$, as in choice (B). Click here to jump to the problem!
 24 Click here to jump to the problem! GR8677 #67 Optics$\Rightarrow$}Polarized light A plane-polarized wave has intensity $I\propto \cos^2\theta$, where $\theta$ is the angle from the wave to the polarization axis. (This is also known as Malus' Law.) An unpolarized wave has intensity $I = const$. Since ETS is generous enough to supply the intensity, one can easily deduce choice (C). Click here to jump to the problem!
 25 Click here to jump to the problem! GR8677 #68 Optics$\Rightarrow$}Aperture Formula The formula that relates the angle of an angular aperture to the wavelength and diameter is $\theta = 1.22 \lambda/d$. Thus, $d=1.22\lambda/\theta$. Plug in numbers to get (C). Click here to jump to the problem!
 26 Click here to jump to the problem! GR8677 #69 Optics$\Rightarrow$}Speed of Light The speed of light is related to the index of refraction by $n=c/v$. Thus, the minimal velocity the particle must have is $v=c/n=2/3c$, since $n=3/2$. Click here to jump to the problem!
 27 Click here to jump to the problem! GR8677 #96 Optics$\Rightarrow$}Interferometer An (effective) path change of $\lambda$ produces a fringe shift. Thus, the interferometer formula is similar to the interference formula at normal incident, $m=2\frac{\Delta L}{\lambda}$. Thus, $m = 2\left( dn/\lambda - d/\lambda \right)=2d/\lambda(n-1)$. Thus, $n=\frac{m\lambda}{2d}+1=1.0002$, as in choice (C). See GR0177.100 on the same site for more info. Click here to jump to the problem!

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...