Chapter 24: Optical Instruments Combinations of Lenses The Camera The Eye The Magnifier The Compound Microscope The Telescope

§24.1 Combinations of Lenses With lenses in combination, the image formed by one lens is the object for the next lens. Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 24, Questions 1, 4, 5, and 14.

The thin lens equation still applies. p is the object distance, q is the image distance, f is the focal length and s is the distance between the lenses. It is possible to have the object for the second lens be virtual (p2<0); here the image formed by the first lens is beyond the second lens.

The magnification of a combination of lenses is just the product of the magnifications for the individual lenses.

Example (text problem 24.2): A converging and diverging lens, separated by a distance of 30.0 cm, are used in combination. The converging lens has f1 = 15.0 cm and the diverging lens has an unknown focal length. An object is placed at 20.0 cm in front of the converging lens; the final image is virtual and is formed 12.0 cm before the diverging lens. What is the focal length of the diverging lens? Given: p1 = 20 cm; f1 = 15 cm. Find that q1 = 60 cm Given: s = 30.0 cm. Find that p2 = -30 cm. Given: q2 = -12 cm and p2 = -30 cm. Find that f2 = -8.6 cm.

§24.2 The Camera Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 24, Questions 6 and 14. A camera forms a real, inverted image. For far away objects, the film must be placed one focal length from the lens.

Now use the thin lens equation Example (text problem 24.17): A person on safari wants to take a photograph of a hippopotamus from a distance of 75.0 m. The animal is 4.00 m long and its image is to be 1.20 cm long on the film. (a) What focal length lens should be used? Part c is not included. Now use the thin lens equation

The object is 2.67 mm long (inverted). Example continued: (b) What would be the size of the image if a lens of 50.0 mm focal length were used? The object is 2.67 mm long (inverted).

§24.3 The Eye The lens is at a fixed distance from the retina (unlike in some cameras where this is adjustable). The lens has a variable focal length, which is adjusted to keep the image distance (q) constant as the object distance (p) varies. Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 24, Questions 2, 3, 8, 15, and 16.

The near point is the closest distance from your eye that an object can be seen clearly. For a normal eye this distance is 25 cm. The far point is farthest distance from your eye that an object can be seen clearly. For a normal eye this distance is . Refractive power of a lens is defined as: where f is the focal length of the lens; typical units of P are diopters (1D = 1 m-1).

For a near sighted (myopic) person, light rays converge before they strike the retina. A diverging lens is placed in the light path. This creates a virtual image closer to your eye than the actual object is.

For a far sighted (hyperopic) person, light rays converge after they strike the retina. A converging lens is placed in the light path. This creates a virtual image farther from your eye than the actual object is.

Given p = 25 cm; q = -1.5m. Here f = +30.0 cm Example (text problem 24.63): If Harry has a near point of 1.5 m, what focal length contact lenses does he require? The near point refers to the closest distance an object can be to see it clearly, in this case 1.5 m. A normal eye has a near point of 25 cm. These corrective lenses must take an object at 25 cm and form a virtual image at a distance of 1.5 m. Given p = 25 cm; q = -1.5m. Here f = +30.0 cm The refractive power of these lenses is (Diopter) Harry is farsighted.

§24.4 The Magnifier The farther an object is from your eye, the smaller it will look. For an object to look bigger the image of it formed on the retina must be made bigger. Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 24, Questions 7 and 12.

Using the triangles in the figure, the angular size  of an object is:

The angular magnification is The largest angular size an object can have and still be seen clearly is when it is placed at your near point. Now the object is placed at the focal point of a converging lens

N is the near point for a person (typically 25 cm) and f is the focal length of the lens used in the magnifier.

Example (text problem 24. 34): An insect that is 5 Example (text problem 24.34): An insect that is 5.00 mm long is placed 10.0 cm from a converging lens with a focal length of 12.0 cm. (a) What is the position of the image? Given: p = 10.0 cm and f = 12.0 cm. Find that q = -60.0 cm; on the same side of the lens as the insect.

(b) What is the size of the image? Example continued: (b) What is the size of the image? (c) Is the image upright or inverted? Since h’>0, the image is upright.

(d) Is the image real or virtual? Example continued: (d) Is the image real or virtual? Since q<0 (the image forms on the same side of the lens as the object), the image is virtual. (e) What is the angular magnification if the lens is close to the eye? Object is at a distance of p, not f from the lens.

§24.5 The Compound Microscope Two converging lenses are used to produce a highly magnified image. The objective lens forms an enlarged real image here. Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 24, Questions 9, 17, and 20. The eyepiece is used to view this image.

The eyepiece is used as a magnifier; the image formed by the objective is placed at the focal point of the eyepiece. The total magnification is the product of the individual magnifications: where L = “tube length” = q0 - f0; N = near point distance; and fobj & feye are focal lengths of the objective and eyepiece respectively.

Example (text problem 24.42): A microscope has an objective lens of focal length 5.00 mm. The objective forms an image 16.5 cm from the lens. The focal length of the eyepiece is 2.80 cm. (a) What is the distance between the lenses? Using figure 24.16, the distance between the lenses is d = feye + qo = 2.80 cm + 16.5 cm = 19.3 cm.

(b) What is the angular magnification? The near point is 25.0 cm. Example continued: (b) What is the angular magnification? The near point is 25.0 cm. (c) How far from the objective should the object be placed? Given: q = 16.5 cm and f = 0.5 cm. Find that p = 0.52 cm.

§24.6 The Telescope A telescope is a combination of lenses and/or mirrors used to collect a large amount of light and bring it to a focus. Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site (www.mhhe.com/grr), Instructor Resources: CPS by eInstruction, Chapter 24, Questions 13, 18, and 19.

A refracting telescope (or refractor) uses lens. The barrel (or tube) length is The angular magnification is

A reflecting telescope uses mirrors (and lenses). Light path of a Cassegrain telescope:

Example (text problem 24.49): A refracting telescope is used to view the moon (diameter of 3474 km & distance from Earth 384,500 km). The focal lengths of the objective and eyepiece are +2.40 m and +16.0 cm, respectively. (a) What should be the distance between the lenses?

(b) What is the diameter of the image produced by the objective? Example continued: (b) What is the diameter of the image produced by the objective? (c) What is the angular magnification?

Summary Combinations of Lens The Camera The Eye The Magnifier The Compound Microscope The Telescope