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Ch.27: Optical Instruments (May 5-7)

Concepts Ch.27

Units Ch.27

Lecture notes May 5,7

Equation Sheet Ch.27

Equation Questions

  1. A question about the total angular magnification of the telescope:
    • the wiki equation has $M_\text{total} = -\frac{f_\text{objective}}{f_\text{eyepiece}}$ but the book has no negative sign. Which is correct? — Ryan Stoner
    • Answer: Traditionally, telescope magnification is always quoted as a positive number. But depending on the configuration of the telescope, the image can be inverted or upright. In the spirit of the previous chapter, we assign the negative sign when the image is inverted, but the book doesn't do that.Prof. Nicholas Kuzma 2015/05/24 22:31
Chapter 27
Any lens or mirror Size of Aperture
$f$-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
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
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
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
$M_\text{total}=m_\text{objective}\!\cdot\! M_\text{eyepiece}$
Length of microscope $L\approx\,d_i+\,f_\text{eyepiece}$
Simple compound telescope
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
    1. tense muscle
    2. the focal length of the eye's lens is shortened (refractive power is the highest)
    3. 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
    1. relaxed muscle
    2. the greatest focal length of the eye's lens (refractive power is the lowest)
    3. 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.
      1. 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-1.
    • 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
    1. 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
    2. 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.


  1. To be able to use diopter units, convert all distances to SI:
    • $d_i=0.025\,$m
    • $N=0.17\,$m
  2. Use the thin lens equation for the eye's lens:
    • $\frac{1}{f}= \frac{1}{d_o} + \frac{1}{d_i}$
  3. Far point:
    • $d_o= \infty$
    • $\frac{1}{d_o}=$ $\frac{1}{\infty}=0$
    • $\frac{1}{f_\text{far}}=$ $\frac{1}{\infty} + \frac{1}{0.025\,{\text m}}=$ $\ 40\,{\text m}^{-1}$ = 40 dpt
    • SI Units: Diopter (dpt)
    • $f_\text{far}=$ $d_i=2.5\,$cm
  4. 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 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.


Problem 5.3

A 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.


Problem 5.4

A simple, 35-mm film camera is used in 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?


  1. 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
  2. Image Size
    • $h_i = h_o\cdot\left(-\frac{d_i}{d_o}\right)$ $=1.7\,{\text m}\cdot\left(-\frac{0.08\,{\text m}}{5\,{\text m}}\right)$ $= -27.2\,{\text{mm}}$
  3. A negative image size means an inverted orientation
  4. 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:

  1. $\times\, 500$
  2. $\times\, 100$
  3. $\times\, 40$

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 $d_i^\text{eyepiece}=\infty$.

  • Can anyone figure this question out? It is included in our lecture notes,but is not making sense.


  1. The approximate formula is $\left|M_\text{total}\right|=\left|m_\text{objective}\right|\cdot\! M_\text{eyepiece}$ $\approx\left(\frac{d_i}{f_\text{objective}}\right)\left(\frac{N}{f_\text{eyepiece}}\right)$
    • We need to find $L$ from $\left|M_\text{total}\right|$ , $f_\text{eyepiece}$ and $f_\text{objective}$:
      • $d_i\approx$ $\left|M_\text{total}\right|\,f_\text{objective}\left(\frac{f_\text{eyepiece}}{N}\right)$
      • $L=d_i+f_\text{eyepiece}$ $\approx f_\text{eyepiece}\left(\left|M_\text{total}\right|\,\frac{f_\text{objective}}{N}+1\right)$
    • Tabulating this expression for $\left|M_\text{total}\right|=500,100,40$ gives:
      • $L_{\times 500}\approx 22$ cm
      • $L_{\times 100}\approx 6$ cm
      • $L_{\times 40}\approx 3.6$ cm
  2. The exact formula is $\left|M_\text{total}\right|=\left|m_\text{objective}\right|\cdot\! M_\text{eyepiece}$ $=\left(\frac{d_i}{d_o}\right)\left(\frac{N}{f_\text{eyepiece}}\right)$
    • Again, we need to find $L$ from $\left|M_\text{total}\right|$ , $f_\text{eyepiece}$ and $f_\text{objective}$:
      • $d_i=$ $\left|M_\text{total}\right|\,d_o\left(\frac{f_\text{eyepiece}}{N}\right)$
      • $d_o=$ $\left(\frac{1}{f_\text{objective}}-\frac{1}{d_i}\right)^{-1}$
    • The two equations above have two unknowns: $d_i$ and $d_o$, we can solve them as a system of equations
      • using variable elimination (e.g. express $d_o$ from one equation and plug it into the other)
      • The final result of solving these equations for $d_i$ and $d_o$:
        • $d_i=$ $f_\text{objective}\left(\left|M_\text{total}\right|\frac{f_\text{eyepiece}}{N}+1\right)$
        • $d_o=$ $f_\text{objective}\left(\frac{N}{f_\text{eyepiece}\left|M_\text{total}\right|}+1\right)$
    • $L=$ $d_i+f_\text{eyepiece}$ $=f_\text{eyepiece}\left(\left|M_\text{total}\right|\,\frac{f_\text{objective}}{N}+1\right)+f_\text{objective}$
    • Tabulating this expression for $\left|M_\text{total}\right|=500,100,40$ gives now:
      • $L_{\times 500}= 22.4$ cm
      • $L_{\times 100}= 6.4$ cm
      • $L_{\times 40}= 4.0$ cm
    • According to our formula, these exact answers are precisely obtained by adding $f_\text{objective}$ to the approximate expression.

Problem 5.6

A simple telescope of length $L=40\,$cm has two options for the eyepiece lens:

  1. $f_\text{eyepiece}=4\,$cm
  2. $f_\text{eyepiece}=1\,$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.

$\;\;$ Using the $L=$ $f_\text{objective}+f_\text{eyepiece}$ formula,

  1. $f_\text{objective}=$ $L-f_\text{eyepiece}$ $=36\,$cm
    • $M=\frac{f_\text{objective}}{f_\text{eyepiece}}$ $=9\!\times$
  2. $f_\text{objective}=$ $L-f_\text{eyepiece}$ $=39\,$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

  1. a far-away object ($d_o=\infty$)
  2. a neighbor's house 30 ft away

Also find the magnitude of angular magnification in each case.


  • To find value of $L$ at $d_o=\infty$
    1. use equation, $L= f_\text{obj} + f_\text{eyepiece}$
      • $L= 20\,{\text{cm}} + 0.5\,{\text{cm}}$ $= 20.5\,{\text{cm}}$
    2. For angular magnification: use equation $M=-\frac{f_\text{obj}}{f_\text{eyepiece}}$
      • $M= -\frac{20\,{\text{cm}}}{0.5\,{\text{cm}}}$ $= -\,40$
  • To find value of $L$ at $d_o=30$ ft
    1. Convert to S.I. units.
      • 30 ft = 9.146 m
    2. Use equation $\frac{1}{f_\text{obj}}= \frac{1}{d_o} + \frac{1}{d_i}$ and solve for $d_i$.
      • $d_i= \big(\frac{1}{f_\text{obj}} - \frac{1}{d_\text{o}}\big)^{-1}$
      • $d_i= 0.204$ m $= 20.4$ cm
    3. $L= f_\text{obj} + f_\text{eyepiece}$ $= 20.4\,{\text{cm}} + 0.5\,{\text{cm}}= 20.9\,{\text{cm}}$
    4. For angular magnification: use the above equation
      • $M= -\frac{20.9\,{\text{cm}}}{0.5\,{\text{cm}}}$ $= -\,40.8$

Homework Questions Ch.27

Problem 27.7

What if the f number in dim light when the pupil has expanded to 6.0 mm?

  • focal length is 1.7 cm
  • pupil diameter is 2.0 mm

I also got the wrong answer for calculating the f number when the diameter is 2.0 mm in bright light.

  • I got 8.2, but the answer was 8.5. I'm not quite sure where I went wrong.

Answer 27.7

  • According to the definition of F-number, it is simply $\frac{f}{D}$.
    • So, $\frac{1.7\,{\text{cm}}}{2\,{\text{mm}}}=8.5$ (don't forget to convert mm to cm or vice versa)

Problem 27.11

Part a. In the HW solutions it has a negative sign in the equation $M=-\frac{f_\text{objective}}{f_\text{eyepiece}}$. Is this because the eye piece is concave? I am just confused why it says to set the magnification equal to the negative of the ratio of the focal lengths.

Answer 27.11

Very good question! Traditionally, telescope magnification is quoted without any regard to the + or $-$ signs, since very few astronomers care if a given star looks upside-down or not. But technically speaking, the simplest telescope design discussed in class always inverts the image (makes it upside-down), so, according to the definition of the magnification, it should be negative. — Prof. Nicholas Kuzma 2015/05/12 23:51

Problem 27.29

  1. 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
  2. I am also having trouble with this problem, I'm getting the final image distance being 6.04 cm but the answer was 12 cm. I used $d_i^{(2)}=\left(\frac{1}{f_2}-\frac{1}{d_o^{(2)}}\right)$. And I got $d_o^{(2)}$ to be $-24.29$ from $d_i^{(1)}$-separation. —Shon Tyler
    • Are you forgetting to invert the difference of inverses as well? Also note that everything propagates right to left here, so the diverging lens is the first one, the converging is the second. If your input numbers are $L=20.3$ cm, $f_1=-5.74$ cm, and $f_2=8.47$ cm, then $d_o^{(2)}$ should be $L-d_i^{(1)}=20.3-(-4.3)=24.6$ cm. I'm getting 12.9 cm for the final answer. — Prof. Nicholas Kuzma 2015/05/10 22:22

Problem 27.45

A person's prescription for her new bifocal glasses calls for a refractive power of -0.400 diopters in the distance-vision part, and a power of 1.85 diopters in the close-vision part.

  • What are the near and far points of this person's uncorrected vision?
  • Assume the glasses are 2.00 cm from the person's eyes, and that the person's near-point distance is 25.0 cm when wearing the glasses.

My steps so far 27.45

I did $1/-0.04$ and got $-250$ cm, so 2.52 m for the far point. I then did $1/56-1/25$ and got $-0.45$ m for the near point. I even had my tutor help me, and we could not get the correct answer.

Answer 27.45

  • You need to do a step-by-step detailed calculation here, since the distance from the glasses to the eye is given.
    • Far point: $\frac{1}{f_\text{far}}=\frac{1}{\infty}+\frac{1}{-({\text{F.P.}}-0.02\,{\text m})}$
      • since the lens takes the image at infinity and brings it to the person's far point away from the eye, as a virtual image
      • Then, ${\text{F.P.}}=-\frac{1}{-0.400}+0.02\,{\text m}=2.52$ m
    • Near point: $\frac{1}{f_\text{near}}=\frac{1}{0.25\,{\text m}-0.02\,{\text m}}+\frac{1}{-({\text{N.P.}}-0.02\,{\text m})}$
      • since the lens takes the image at 25 cm (from the eye, 23 cm from the lens) and brings it to the person's near point away from the eye, as a virtual image
      • Solve this for N.P.

Prof. Nicholas Kuzma 2015/05/13 00:21

Problem 27.60

The medium-power objective lens in a laboratory microscope has a focal length $f_\text{objective} = 4.00$ mm. If this lens produces a lateral magnification of $-40.0$, what is its “working distance”; that is, what is the distance from the objective to the objective lens?

  • Q: If the objective lens is used as a magnifying glass, shouldn't the magnification equation be $M=\frac{N}{f}$? The first step in solving this equation in the HW solutions instead uses $m=\frac{d_i}{d_o}$ as the magnification formula, which is used to solve for distance of the object ($d_o$).
  • A: The way we formulate the angular magnification of a microscope is this: the angular size of the final image (as seen by the eye) divided by the angular size of the object held at the near point of the same eye without any optical aid.
    • For the eyepiece, we do use the lens formula. That lens formula assumes some object size (linear size, not angular) which is placed near the focal point of the eyepiece. In a microscope, that physical linear size is that of the first image. Therefore we need to use a linear magnification (not angular) for the objective, to convert the actual object size to the first image size. That is what the HW solution does.

Equation use question

Magnifying Glass - Magnification

chapter_27.txt · Last modified: 2015/05/26 06:24 by wikimanager