exam_2_review

* This was given as a midterm (Exam 2) in 2014 by Prof. Kuzma*.

Solutions are available on D2L.

2014 Physics 203 T-10 | Exam 2 - Midterm | Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ |

- PLEASE PRINT YOUR NAME ABOVE!!!
- Please put away everything except:
- this exam and your Scantron form (use back sides of these pages for drafts you don’t want to be graded).
- one double-sided sheet of notes (formulas, etc.) that you prepared for this exam
- calculator (must have no connection to internet and no text/picture storage)
- pen/pencil/eraser
- clock/watch (no connection to internet, no memory functions)

- Please make sure you are not sitting next to someone else, unless there is no such seats available.
- No exchange of information between students is allowed during the exam!
- Please do not open this booklet until you are instructed to do so. Please do not unstaple this booklet!
- Out of 4 problems, please cross one out, and attempt the other 3.
**New in 2015**: Failure to cross out a problem will cause the grader to cross one out for you randomly.

- For full credit, show your work!
- If you need to take a quick bathroom break:
- please leave this booklet with one of the proctors on your way out
- pick it up on your way back in.

- If you have a question:
- please read the problem carefully again
- If you think some information is missing, please walk down to a proctor with your question.

- Please hand in your finished exam to one of the proctors on your way out.

- speed of light $c$:
- $c_{\text{vacuum}} = 3 \times 10^8$ m/s

- $\epsilon_0 = 8.85 \times 10^{-12}$ F/m
- $\mu_0 = 4\pi \times 10^{-7}\,\frac{\text N}{{\text A}^2}$
- $1^\circ=\frac{\pi}{180}\,{\text{rad}}$

Part of the exam | Score | …out of maximum |
---|---|---|

Question 1 | 2.5 | |

Question 2 | 2.5 | |

Question 3 | 2.5 | |

Question 4 | 2.5 | |

Question 5 | 2.5 | |

Question 6 | 2.5 | |

Question 7 | 2.5 | |

Question 8 | 2.5 | |

Question 9 | 2.5 | |

Question 10 | 2.5 | |

Problem 1 | 25 | |

Problem 2 | 25 | |

Problem 3 | 25 | |

Problem 4 | 25 | |

Total: | 100 |

Which statement below **IS NOT** true for linearly-polarized light of a certain wavelength?

- [….] A) The electric field is zero at the same points in space and time where the magnetic field is zero
- [….] B) The direction of light propagation is perpendicular to both the electric and the magnetic field vectors
- [….] C) The energy density of the magnetic field is equal to that of the electric field at any given point
- [….] D) The electric field magnitude $|\mathbf{E}|$ (the length of the $\mathbf{E}$ vector) is constant both in space and in time, but the direction of $\mathbf{E}$ is not
- [….] E) The intensity of such light, transmitted at normal incidence through a linear polarizer, can be reduced to zero by aligning the polarizer axis in a specific direction

A linearly polarized light beam is traveling along the *x* axis, perpendicular to the plane of a linear polarizer. The angle between the incoming beam's electric field and the polarizer's polarization axis is $\theta = 30^\circ$. What is the transmitted beam's average intensity if the initial beam's average intensity was $1\,\frac{\text W}{{\text m}^2}$?

- [….] A) $\;\;0.024\,\frac{\text W}{{\text m}^2}$
- [….] B) $\;\;0.25\,\frac{\text W}{{\text m}^2}$
- [….] C) $\;\;0.5\,\frac{\text W}{{\text m}^2}$
- [….] D) $\;\;0.75\,\frac{\text W}{{\text m}^2}$
- [….] E) $\;\;0.866\,\frac{\text W}{{\text m}^2}$

Which of the following will **shorten** the focal distance of a concave mirror?

- [….] A) Doubling all of its dimensions (i.e. making a scaled-up version, twice as large as the original)
- [….] B) Making it more curved by reducing its radius of curvature by 10%
- [….] C) Submersing the mirror in water (refraction index $n = 1.333$)
- [….] D) Replacing the red right with blue light
- [….] E) Removing the outer margin of the mirror, leaving only a small circular area near the center

A candle is burning at the center of a large, perfectly-spherical glass chamber, coated with silver on the outside. Where (if at all) will all the light emitted by the candle be focused?

- [….] A) At the candle itself
- [….] B) At the wall of the chamber
- [….] C) At a distance half way between the candle and the wall of the chamber
- [….] D) The light will not focus anywhere
- [….] E) Outside of the chamber

Professor, I believe the explanation is probably pretty simple, but what is the reasoning behind the correct answer(A)? —Nichole K.

Two equivalent ways:

- Tracing each ray: any ray from the candle strikes the mirror at a 90° angle, reflects right back at the candle
- Using the $1/f=1/d_i+1/d_o$ formula, noting that $f=R/2$ and $d_o=R$

A thin convex lens projects a bright object, located 28 mm from the lens, onto a screen 2 m away in sharp detail. What is the focal distance of this lens?

- [….] A) $\;\;2$ m
- [….] B) $\;\;1.87$ m
- [….] C) $\;\;27.6$ mm
- [….] D) $\;\;-28.4$ mm
- [….] E) $\;\;36.21$ m

The angular distance between (i.e. the angle between the light beams emitted by) Alcor and Mizar in the constellation of Big Dipper is 0.197°. At what angle will these objects appear relative to each other if viewed through an amateur telescope, in which the focal distance of the objective is 60.0 cm and the focal distance of the an eye-piece is 1.50 cm?

- [….] A) $\;\;0.197^\circ$
- [….] B) $\;\;0.0049^\circ$
- [….] C) $\;\;2.63^\circ$
- [….] D) $\;\;11.82^\circ$
- [….] E) $\;\;7.88^\circ$

A student with a near point $N = 15$ cm is using a convex lens of focal distance $f = 1.5$ cm as a magnifying glass. Approximately, what is the smallest object this student can resolve through this magnifier? Assume that without one, the student can resolve objects down to $50 \,\mu$m.

- [….] A) $\;\;50\,\mu$m
- [….] B) $\;\;500\,\mu$m
- [….] C) $\;\;5\,\mu$m
- [….] D) $\;\;33\,\mu$m
- [….] E) $\;\;\infty$

Two thin lenses with focal distances of 10 cm and 25 cm are placed, touching, as shown. What is the focal distance of this combination?

(*Hint*: what is the refractive power of a combination in terms of the refractive power of the individual components?)

- [….] A) $\;\;35$ cm
- [….] B) $\;\;7.14$ cm
- [….] C) $\;\;2.5$ m
- [….] D) $\;\;15$ cm
- [….] E) $\;\;14$ cm

What optical instruments can one make using two identical convex lenses, each $\,f = + 1$ cm ?

- [….] A) Corrective glasses for a nearsighted person
- [….] B) Corrective glasses for a farsighted person
- [….] C) A telescope
- [….] D) A microscope
- [….] E) A system for converting a mirror image to a normal image (i.e. Я to R)

What happens if a student with normal vision tries on a pair of eyeglasses prescribed for somebody else who is quite nearsighted?

(Hint: think in terms of refractive power)

- [….] A) The curious student will be like a nearsighted person without glasses: unable to see far objects
- [….] B) (S)he will experience “farsightedness” — will not be able to focus at her/his usual near point
- [….] C) Will finally be able to discern very small print, using the glasses as magnifiers
- [….] D) Will be in danger of suffering eye damage on a sunny day, as the glasses tend to focus the sunlight
- [….] E) Will be very confused, as everything will appear upside-down

A linearly polarized light beam of average intensity $1000\,\frac{\text W}{{\text m}^2}$ propagates along the $x$ axis. Its electric field direction is along the $y$ axis. The beam passes through a number of polarizing plates (one after another), at normal incidence.

- A) [4 points] If the angle of the 1
^{st}polarizer's axis with respect to the y axis is $\theta_1 = 15^\circ$, what is the average transmitted intensity immediately after this polarizer? - B) [3 points] By what angle $\theta_1$ should one orient the 1
^{st}polarizer's axis with respect to the $y$ axis, in order to prevent any light from propagating through the rest of the optical system? - C) [4 points] If only $\frac{1}{4}$ of the original intensity propagates through the first polarizer, what is the angle $\theta_1$ between that polarizer's axis and the $y$ axis?
- D) [3 points] A 2
^{nd}polarizer, with the axis at an angle $\theta_2$ from the $y$ axis, is placed after the 1^{st}polarizer. What should the difference $\theta_2 - \theta_1$ be, to prevent any light from passing through this system? - E) [4 points] Find the transmitted intensity through the whole system if $\theta_1 = 15^\circ$ and $\theta_2 = 30^\circ$.
- F) [5 points] Now 6 polarizers, whose axes are uniformly covering the range from $\theta_1 = 15^\circ$ to $\theta_6 = 90^\circ$, are defined by $\theta_n = n \cdot 15^\circ$, with $n = 1, 2, 3, 4, 5, 6$. (E.g. $\theta_2 = 30^\circ$, $\theta_3 = 45^\circ$, etc.) Find the final transmitted intensity (3.5 pts) and the polarization direction (1.5 pts) of the final output beam.
- G) [2 points] Assuming perfect polarizers (no losses beyond the $\cos 2\theta$ formula), calculate the final intensity and polarization direction of the beam passing through a system of 60 polarizers, uniformly spanning the range from $\theta_1 = 1.5^\circ$ to $\theta_{60} = 90^\circ$. Here $\theta_n = n \cdot 1.5^\circ$, with $n = 1, 2, 3, \ldots 60$.

**Q**: For parts F) and G), I do not understand how to determine direction of the polarization. Could you clarify please. Thanks —*Ryan Stoner***A**: After normally passing through a perfect linear polarizer, the exiting light is always linearly polarized in the direction of that polarizer's axis. Therefore, for a system of several such polarizers, you just need to figure out the direction of the last polarizer's axis.

—*Prof. Nicholas Kuzma 2015/05/11 21:57*

Two identical convex lenses of focal distance $f = +10$ cm are arranged as shown along the $x$ axis. The lenses are located at $x = 20$ cm and 60 cm. Initially, a 2-cm tall object is at $x_o = 0$.

- A) [6 points] Find the $x$ coordinate, magnification, and height of the 1
^{st}image (formed by the first lens) - B) [3 points] Is this 1
^{st}image real or virtual? Upright or inverted? Draw it (as an arrow) on the figure above. Does the $x$ coordinate found in A) make sense on this drawing? - C) [5 points] Find the $x$ coordinate and height of the final image formed by this optical system. Is it virtual?
- D) [1 point] Draw the final image as an arrow on the figure above. Does the $x$ coordinate found in C) make sense on this drawing?
- E) [4 points] The object is now moved to the right, to a new location $x_o' = + 2$ cm along the $x$ axis. Find the $x$ coordinate of the 1
^{st}image in this case. - F) [4 points] Find the $x$ coordinate of the final image (after both lenses) of the displaced object. Did the final image move in the same direction as the object? Also compare the displacement distance $\Delta x$ (i.e. change of the $x$ coordinate) of the final image to that of the object.
- G) [2 points] Describe in words (or draw) what happens if the object moves to a location $x_o'' = +10$ cm.

*Hint*: assume the object is a point source (of zero height), and just follow the rays.

A farsighted person cannot focus closer than $N = 60$ cm. Assume the distance from the person's eyes' lens to the retina is $d_\text{eye} = 2.4$ cm. A 1-cm tall object is initially 60 cm away from the eye.

- A) [3 points] At what angle $\theta_\text{without}$ does the person see this object without any optical aids?

*Hint*: you can assume that for this small angle $\tan\theta \approx \sin\theta \approx \theta$ (in radians!) - B) [4 points] What is the size of the image formed on the retina?
- C) [4 points] What is the focal distance (and its inverse, refractive power) of the eye at the near point?
- D) [5 points] Find the focal distance of a corrective lens that would just enable this person to see the same object sharply at a distance of 20 cm. Is it a converging or diverging lens?

*Hint*: is the image formed by the corrective lens real or virtual? Where is it seen by the eye? - E) [4 points] With the corrective lens, what is the size of the final image of the object on the retina?

*Hint*: assume the corrective lens is very close to the eye.

*Hint #2*: how does bringing the object from 60 cm to 20 cm change its angular size? - F) [3 points] Now consider the corrective lens to be a magnifier glass. Find its angular magnification (for the image formed at the eye's near point $N$).
- G) [2 points] Compare the previous answer to the ratio of the answers in parts E) and B). Discuss.

- (Hi Professor Kuzma, I don't understand why we use negative 60cm for Distance of the object? Is it because our image is gonna be virtual on the same side of the lens? —
*Mahnoushrafat*- No, the object is now at 20 cm. The lens is projecting it at 60 cm away from the eye. So, the image is at 60 cm. And because this image is behind the lens, it is virtual. —
*Prof. Nicholas Kuzma 2015/05/14 02:29*

A simple telescope has an objective of a focal distance 59 cm and an eyepiece of a focal distance 1 cm. It is designed for maximum comfort of star gazing, forming images at $\infty$.

- A) [3 points] What is the length of this telescope (lens-to-lens distance)?

- B) [4 points] What is its total angular magnification? (at this point, the sign is not important, just the number)

**In parts C) – G) below**, this telescope is used, *without any adjustments*, to look at a house 600 m away.
Keep all calculations to 5 significant digits and *do not* use any approximations.

- C) [5 points] Find how far from the objective will the 1
^{st}image of the house form (i.e., $d_i$ of the objective).

- D) [2 points] Is this image upright or inverted? Real or virtual?

- E) [6 points] This image is the object for the eyepiece. Where is the 2
^{nd}image of the house formed by the eyepiece, relative to the eyepiece?

- F) [2 points] Is it a real or virtual image? Does the house appear upside-down to the viewer?

- G) [3 points] If the viewer's near point is $N = 15$ cm, will he or she be able to see this image sharply? Will it be a comfortable experience without adjusting the length of the telescope?

exam_2_review.txt · Last modified: 2015/05/14 09:31 by wikimanager

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