Chapter 25
The Giancoli onLine Tutor

Optical Instruments

I. Cameras

A simple camera is shown at the left. The camera works by focusing an image on the film at the right. The governing equation is shown at the right. For distant objects 1/do ~ 0 and the image distance will be at the focal length (normally 50 mm). As the object moves closer the image distance increase and you must focus by moving the lens further away from the film. A telephoto lens used for a distant object works by changing the focal length. Since the lateral magnification is given by di/do, which closely equals f/do, changing from a focal length of 50 mm to 200 mm will increase the images size on the film by a factor of 4. Conversely, a wide angle lens works by decreasing the focal length of the lens.

The diaphragm controls the size of the lens opening. The smaller the size of the opening, the more the ability the lens has to 'focus' on objects at different distances. The lens opening is described by the f-stop which is equal to f/D where D is the diameter of the lens in mm. A 25 mm opening would be recorded as f/2 (50mm/2 = 25 mm). The standard openings are: f/2.8, f/4, f/5.6, f/8, f/11, f/16, f/22 and f/32. As you go up the list you decrease the radius by SQRT(2) and hence the emitting area by a factor of 2. Thus each step decreases the light entering the camera and reaching the film by a factor of 2. To compensate for this you have to double the shutter exposure time.

Interactive Examples


Review the above material and answer the following questions.

  1. If a camera has a focal length of 50 mm and focuses clearly objects at infinity ( that is di = f ), how far would the lens have to be moved to the nearest mm to focus on an object 1 m away.
    Enter your answer and click outside.

  2. If the diaphragm of the shutter is set to 10 mm, what is its f-stop?
    Enter your answer and click outside.


II. Eyeglasses

A farsighted eye is unable to see objects up close (25 cm). The nearpoint, np, (the closest point at which the eye can focus) is larger than 25 cm. To correct this you use an eye glass that images an object held at 25 cm at the actual nearpoint (-np). The minus sign occurs because the image must be on the same side of the lens as the object, that is, a virtual image.

-1/np + 1/25= 1/f


f will be the needed focal length (always positive for this condition) expressed in centmeters. The lens power, P, is 100/f where f is in centimeters .

Interactive Example


If the nearpoint is 1.5 meters, the lens power is...
Enter your answer and click outside.

The above calculation assumes that there is negligible distance between the eye and the eye piece. If the eyeglass was 2 cm in front of the eye then the object distance to use would have been 23 cm.

A nearsighted eye can not focus on distant objects (infinity); it has a far point,fp, of less than this. An object at infinity must be (virtually) imaged at -fp. Since 1/infinity = 0, it follows that

-1/fp = 1/f and P = -100/fp


Interactive Example


If the farpoint of an eye is 1.5 m, the lens power is...
Enter your answer and click outside .

if the eyeglasses are 2 cm in front of the eye, you need to subtract 2 cm from fp.

III. Magnifying Glasses

A simple optical instrument is the magnifying glass, a simple converging lens held so that the object is just inside f, for maximum viewing (angular ) magnification, M,

M= 1+25/f (f in cm),

or at the focal point f (for easy viewing) and a magnification given by

M= 25/f


Interactive Example


What is the maximum magnification of a magnifying glass with a focal length of 5 cm?
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IV. Telescopes

A refracting (astronomical) telescope utilizes an objective lens with a long focal length, fo, and an eye ptece with a short focal length, fe. the length,L of the telescope is very close to equaling fo + fe. The magnification is

M = - fo/fe


Interactive Example


What is the magnification of a 160 cm long telescope with an eyepiece with fe = 5 cm?
(ignore the minus sign, it just means the image is inverted)
Enter your answer and click outside.

IV. Microscopes

The microscope (length l) uses an objective lens to form a real laterally magnified, m, image at or just inside the focal point of the eye piece. The magnification, m, is given by

m = di/do= (l-de)/do (minus sign ignored)



The eye piece gives an angular magnification , M e= 25/f, for an overall magnification:

M = (25/f) (l-de)/do



Interactive Examples


  1. A 25 cm microscope has an objective with 40X magnification and a 10x eyepiece (focused for ease of viewing), what is the overall magnification?
    Enter your answer and click outside .
  2. What is fe? Enter your answer and click outside.

  3. What is the object distance?
    Enter your answer and click outside .

  4. What is fo?
    Enter your answer and click outside.

V. Optical Resolution

V-a. Telescopes

Any light entering a telescope or microscope is going through a circular aperture , and like light going througe a slit, a diffraction pattern will result. When viewing two close together 'point' stars, both will turn into circular disks surrounded by rings. As shown in the figure, as the stars move closer together, their diffraction pattern begin to overlap. If they become much closer then, as shown at the bottom of the figure, they will become indistinguishable. The generally accepted minimum angular separation that is resolvable is given by

q = 1.22l/D,

where 1.22 results because the opening is a circle instead of a slit, D is the diameter of the lens and l the wavelength of the light.
For earth based optical telescopes atmospheric disturbances put an even greater limitation on resolvability. The above formula can best be thought of as the intrinsic resolution.

Interactive Example


What must be the diameter of a radio telescope using .05 m radio waves if it is to have the same intrinsic resolution as an optical telescope with a 10 inch (25.4 cm) objective using 500 nm light?
Enter your answer and click outside .


V-b. Microscopes


For microscopes we usually use the resolving power, RP, in meters instead of angle and it is given by

RP = 1.22lfo/D


where fo is the focal length of the objective lens and D its diameter. The RP gives you the minimum separation between two points that can be resolved

Interactive Example


What is the resolving power of a microscope in nm with an objective D = 1.2 cm and fo equal to 0.4 cm using light of 600 nm?.
Enter your answer and click outside .

finis
© 1999 Carl Adler mailto:Carl@Image-ination.com