chapter_27

# Ch.27: Optical Instruments (May 6-8) ## Lecture notes May 6,8

• May 6: notes10corr.pdf - please download corrected notes! (On page 6, $\;f=...\;$ is replaced with $\;\frac{1}{f}=...\;$)

## 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 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 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 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 $\infty$) $L=f_\text{objective}+\,f_\text{eyepiece}$

• 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.

Steps:

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. Steps:

### 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. Steps:

### 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? Steps:

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

### 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.

Steps:
$\;\;$ 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. Steps:

## Homework Questions Ch.27

### Problem 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 2014/05/11 09:03 