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===Quick links=== ==This wiki== * [[start|Main classwiki page]] * [[CLASS INFORMATION|Class Info]] * {{:Kuzma_Syllab_Spr2014_v4.pdf|Syllabus}} * [[FREQUENTLY ASKED QUESTIONS|FAQ]] * [[CLASS MATERIALS|Class materials]] * [[PHYSICS LABORATORY|Labs]] * [[|Schedule]] * [[PHYSICS WORKSHOP|Workshop]] * {{:workshops:ph299_syllabus_14sp.pdf|W/S syllabus}} * [[Computational Projects|Projects]] * [[White noise project|White noise]] * [[Rainbow project|Rainbow]] * [[Digital sound project|Digital sound]] * [[Announcements]] ==Earlier material== * [[Chapter 13]] * [[Chapter 14]] * [[Exam 1 review]] * [[Chapter 25]] * [[Chapter 26]] * [[Chapter 27]] * [[Exam 2 review]] * [[Final exam review]] ==Previous wikis== * [[|Ph202 - 2014]] ==Other learning tools== * [[|University D2L site]] * [[|Text & homework]] \\ <sub><color magenta>PH203KUZMASPRING2014</color></sub> ==Knowledge & computation== * [[|Wolfram]] $\alpha$ * [[wp>Physics_portal|Wikipedia]] * [[|Physical constants]] * [[| The Physics Hypertextbook]] * [[| HyperPhysics]] ==Add more by editing:== * [[sidebar|This sidebar]] * [[Tasks to do]] ==Help for editors== * [[doku>wiki:syntax|Help on wiki codes]] * [[|Help on wiki math]] * [[Tips on editing]] =="Sandboxes" for practice== * [[Draft page|Practice here]] * [[Draft page 2|Or here if locked-out]]

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======Ch.27: Optical Instruments (May 6-8)====== {{ :images:eye.png?nolink|}} =====Concepts Ch.27===== * [[wp>Optical power|Refractive power]] * [[wp>Human eye]] * [[wp>Cornea]] * [[wp>Iris (anatomy)|Iris]] * [[wp>Lens (anatomy)|Lens]] * [[wp>Ciliary muscle]]s * [[wp>Aqueous humor]] * [[wp>Retina]] * [[wp>Rod cell|Rods]] * [[wp>Cone cell|Cones]] * [[wp>Blind spot (vision)|Blind spot]] * [[wp>Optic nerve]] * [[wp>Accommodation (eye)|Accommodation]] * [[wp>Presbyopia#Focusing_mechanism_of_the_eye|Near point]] * [[wp>Far point]] * [[wp>Amplitude of accommodation|Power of accommodation]] * [[wp>Myopia|Nearsightedness]] * [[wp>Hyperopia|Farsightedness]] {{ :images:canon.png?400|}} * [[wp>Camera]] * [[wp>Magnifying glass]] * Angular magnification * [[wp>f-number]] * [[wp>Optical microscope]] * [[wp>Optical telescope]] * [[wp>Spherical aberration|Spherical]] and [[wp>Chromatic aberration]]s * [[wp>Achromatic lens]] * [[wp>Apochromat|Apochromatic lens]] * [[wp>Superachromat|Superachromatic lens]] =====Units Ch.27===== * [[wp>Diopter]] (dpt) = 1 m<sup>-1</sup> * [[wp>Minute of arc|(Arc)minute]] = (1/60)$^\circ$ * (Arc)second = (1/3600)$^\circ$ =====Lecture notes May 6,8===== * May 6: {{:notes:notes10corr.pdf|}} - <color red>please download corrected notes!</color> (On page 6, $\;f=...\;$ is replaced with $\;\frac{1}{f}=...\;$) * May 8: {{:notes:notes11s.pdf|}} - please download before class =====Equation Sheet Ch.27===== ^ Chapter 27 ||| | Any lens or mirror | Size of Aperture\\ $f$-number\\ (dimentionless) | $f$-number$=\frac{\text {Focal length}}{\text {diameter of aperture}}=\frac{f}{D}$ | | Any lens or mirror | Refractive power\\ Ability to refract light\\ SI Unit: ${\text m}^{-1}=\,$dpt | \\ Refractive power = $\frac{1}{f}$ | | Magnifying glass\\ {{ :figs:eq27fig1.png?nolink |}} | Angular magnification\\ of the magnifying glass | $M=$ $\frac{\theta_\text{with}}{\theta_\text{without}}$ $\approx\frac{\;\frac{h_o}{f}\;}{\frac{h_o}{N}}=\frac {N}{f}$ | |:::| Angle without a magnifier\\ //N// = near point of a person | $\theta_\text{without}=\frac{h_o}{N}$ | |:::|angle with magnifier | $\theta_\text{with}$ $=\frac{h_o}{d_o}$ $\approx\frac{h_o}{f}$ | |:::| image at infinity | $M=$ $\frac{N}{f};\;\;\;\;$ $m=\infty$ | |:::| image at a person's near point | $M=m=$ $1 + \frac{N}{f}$ | | Simple compound microscope \\ {{ :figs:eq27fig2.png?nolink |}} | The magnification\\ produced by the objective\\ $d_o \approx f_\text {objective}$ | $m_\text {objective}=-\frac{d_i}{d_o}$ $\approx -\frac{d_i}{f_\text {objective}}$ | |:::| Angular magnification of\\ the eyepiece | $M_\text {eyepiece}=\frac{N}{f_\text {eyepiece}}$ | |:::| Total angular magnification of\\ the microscope: ($-$) sign\\ means an inverted image | $M_\text{total}=m_\text{objective}\!\cdot\! M_\text{eyepiece}$\\ $=\left(-\frac{d_i}{f_\text{objective}}\right)\left(\frac{N}{f_\text{eyepiece}}\right)$ | |:::| Length of microscope | $L\approx\,d_i+\,f_\text{eyepiece}$ | | Simple compound telescope \\ {{ :figs:eq27fig3.png?nolink |}} |Total angular magnification\\ of the telescope | $M_\text{total}=\frac{\theta'}{\theta}$ $=-\frac{f_\text{objective}}{f_\text{eyepiece}}$ | |:::| Object's angular size | $\theta=-\frac{h_i}{f_\text{objective}}$ | |:::| Final image angular size | $\theta'=\frac{h_i}{f_\text{eyepiece}}$ | |:::| Length of telescope (focus on $\infty$) | $L=f_\text{objective}+\,f_\text{eyepiece}$ | ====Additional notes==== * Viewing a nearby object - tense muscle - the focal length of the eye's lens is shortened (refractive power is the highest) - a farsighted (hyperopic) person will see a fuzzy image of an object between the eye and the near point due to a lack of refractive power that causes the image of the object to form behind the retina. * A converging lens can correct this by increasing the overall refractive power, shortening the focal length to focus on the retina. * Viewing a distant object - relaxed muscle - the greatest focal length of the eye's lens (refractive power is the lowest) - a nearsighted (myopic) person will see a fuzzy image of a distant object beyond the far point due to having too much refractive power which causes the image of the object to form short of the retina. - A diverging (concave) lens can correct this by reducing the overall refractive power, so that the image forms exactly on the retina. * Refractive power * Defined as $\frac{1}{f}.$ * Units: diopter = m<sup>-1</sup>. * The shorter the focal length, the more strongly a lens refracts the light (thus the higher refractive power). * Power of Accommodation: * The difference in the refractive power of the eye at its near and its far points. * Magnifying glass - Nothing more than a simple convex lens, which magnifies by allowing to focus on an object that is closer than the near point of the eye - The larger the image that is formed by the object on the retina, the larger it will appear =====Examples Ch.27===== ====Problem 5.1==== Assuming the depth of the eye is 2.5 cm, find the focal distance for your eye's lens necessary to focus either at the far point ($\infty$) or at the near point //N// = 17 cm. Steps: - To be able to use diopter units, convert all distances to SI: {{ :figs:lec10f12.png?nolink|}} * $d_i=0.025\,$m * $N=0.17\,$m - Use the thin lens equation for the eye's lens: * $\frac{1}{f}= \frac{1}{d_o} + \frac{1}{d_i}$ - Far point: * $d_o= \infty$ \\ <color white>.</color> * $\frac{1}{d_o}=$ $\frac{1}{\infty}=0$ \\ <color white>.</color> * $\frac{1}{f_\text{far}}=$ $\frac{1}{\infty} + \frac{1}{0.025\,{\text m}}=$ $\ 40\,{\text m}^{-1}$ = 40 dpt \\ <color white>.</color> * SI Units: Diopter (dpt) \\ <color white>.</color> * $f_\text{far}=$ $d_i=2.5\,$cm \\ <color white>.</color> - Near point: * $\frac{1}{f_\text{near}}=$ $\frac{1}{0.17\,{\text m}}+\frac{1}{0.025\,{\text m}}= 45.9\,{\text m}^{-1}$ = 45.9 dpt * SI Units: Diopter (dpt) * $f_\text{near}=$ 2.18 cm ---- ====Problem 5.2==== A [[wp>Hyperopia|farsighted]] person's near point (the closest point of sharp focus) is at $N>20$ cm. Calculate, approximately (using refractive power approximation) and precisely, what corrective lens should be placed 2 cm in front of the person's eye to enable sharp focus at 20 cm. Consider the following cases of farsightedness: $N=2\,$m, $1\,$m, $60\,$cm, $40\,$cm, $30\,$cm, $25\,$cm. {{ :figs:lec10f21.png?nolink |}} Steps: - * ---- ====Problem 5.3==== A [[wp>Miopia|nearsighted]] person's far point (the farthest point of sharp focus) is at $X$. Calculate, approximately (using refractive power approximation) and precisely, what corrective lens should be placed 2 cm in front of the person's eye to enable sharp focus far away (at $\infty$). Consider the following cases of nearsightedness: $X=4\,$m, $2\,$m, $1\,$m, $50\,$cm, $30\,$cm, $20\,$cm. {{ :figs:lec10f31.png?nolink |}} Steps: - * ---- ====Problem 5.4==== A simple, 35-mm film camera is used in [[wp>Landscape format|landscape mode]] to take a picture of a 1.7-m tall person from a distance $d_o=5\,$m. Assuming the camera's only thin lens is $d_i=8\,$cm from the film, find the focal distance of the lens required for this image. Also, find the image size and orientation. Will it fit on the 35 mm film? {{ :figs:lec10f41.png?nolink |}} Steps: - Focal length * $\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$ * $f = \left(\frac{1}{5\,{\text m}} + \frac{1}{0.08\,{\text m}}\right)^{-1}$ = 7.8 cm - Image Size * $h_i = h_o\cdot\left(-\frac{d_i}{d_o}\right)$ $=1.7\,{\text m}\cdot\left(-\frac{0.08\,{\text m}}{5\,{\text m}}\right)$ $= -27.2\,{\text{mm}}$ - A **negative** image size means an **inverted** orientation - Since the magnitude of the image size $\big|d_i\big|=27.2\,{\text{mm}}$ $<35\,{\text{mm}}$, it will fit on the film ---- ====Problem 5.5==== A microscope that consists of the objective with the focal distance $f_\text{objective}=4\,$mm and an eyepiece $\big(\,f_\text{eyepiece}=2\,{\text{cm}}\big)$, is used by a person with a near point $N=20\,$cm at the following (angular) magnifications: - $\times\, 500$ - $\times\, 100$ - $\times\, 40$ Find, using an approximate formula for microscope magnification, as well as using a precise calculation, the distance $L$ between the two pieces of optics necessary for each of the above magnifications, assuming the most comfortable viewing distance of the final image $d_i^\text{eyepiece}=\infty$. {{ :figs:lec11f12.png?nolink|}} Steps: - * ---- ====Problem 5.6==== {{ :figs:lec11f21.png?nolink|}} A simple telescope of length $L=40\,$cm has two options for the eyepiece lens: - $f_\text{eyepiece}=4\,$cm - $f_\text{eyepiece}=1\,$cm Calculate the focal distance of the objective lens necessary for each option, as well as the (magnitude of the total angular) magnification in each case. Steps:\\ $\;\;$ Using the $L=$ $f_\text{objective}+f_\text{eyepiece}$ formula, - $f_\text{objective}=$ $L-f_\text{eyepiece}$ $=36\,$cm * $M=\frac{f_\text{objective}}{f_\text{eyepiece}}$ $=9\!\times$ - $f_\text{objective}=$ $L-f_\text{eyepiece}$ $=39\,$cm * $M=\frac{f_\text{objective}}{f_\text{eyepiece}}$ $=39\!\times$ ---- ====Problem 5.7==== Binoculars using a simple telescope design have $\;f_\text{objective}\!=\!20\,$cm and $\;f_\text{eyepiece}\!=\!0.5\,$cm. The distance $L$ between the two lenses can be adjusted to focus on a specific object at a distance $d_o$. Find the value of $L$ necessary to focus on - a far-away object ($d_o=\infty$) - a neighbor's house 30 ft away Also find the magnitude of angular magnification in each case. {{ :figs:lec11f31.png?nolink |}} Steps: - * =====Homework Questions Ch.27===== ---- ====Problem 29==== {{ :figs:hw27_29.jpg?nolink |}} - Where am I wrong in this calculation? * I think you forgot to take an inverse of $\left(\frac{1}{-5.76}-\frac{1}{18.5}\right)$ . Your number, $-0.22$ cm, is awfully short for such a system: it should be much closer to $-5.76$ cm, about $-4$ or $-5$, since $\frac{1}{18.5}$ is a relatively small correction (~30%) to $\frac{1}{-5.76}$. --- //[[|Nicholas Kuzma]] 2014/05/11 09:03// ----

chapter_27.txt · Last modified: 2014/05/18 19:24 by wikimanager