/6 Due Monday, 8 December 2014
LM01. Your prof’s corrective lenses sit about 1.00 cm in front of each eye’s lens. Since his eyeball is elongated (the distance between his biological lens and his retina is greater than 2.45 cm), his biological lenses focus images of far away things to a point in front of his retina; thus, he needs diverging glasses lenses to correct for this. Diverging lenses diverge the rays so that in the end they focus further back in the eyeball where his retina is. Assume that the normal range of focal lengths of the human eye is 2.2313 – 2.45 cm and that the distance from your prof’s eye’s lens to his retina is 2.8755 cm to answer the following.
a) Calculate the required focal length of his corrective lenses so that he can clearly see things that are 25 to infinity cm in front of his eyes. (Toughie!! Be careful when you write the expression for the second object distance. Remember that the object for the eye’s lens is treated as real and that the corrective lens’s f is < 0.)
b) Draw two principal ray diagrams; one showing an object 25 cm in front of his eye and his corrective lens’s image, the second showing this first image (the eye’s lens’s object) and the final image.
c) Is the final image real or virtual? Upright or inverted? What is the overall magnification?
LM03. (Toughie!!!) An object is placed 15 cm in front of a converging lens with f = 10.0 cm. 20.0 cm behind the lens is a diverging mirror with a radius of 28 cm. Find the location of the final image and determine its nature (real or virtual), whether it is erect or inverted, and its magnification. Draw a principal ray diagram for each step.
LM04. A particular microscope has an objective lens with f = 5 cm and an eyepiece with f = 5 cm. In this configuration, the object (an amoeba that is roughly spherical with a diameter of 0.1 mm) is 10 cm in front of the objective lens, and the eyepiece is 14.5 cm on the other side of the objective. Your eye is 1 cm from the eyepiece (see diagram). (NOTE: Occasionally you see lenses represented by vertical double arrows as in the diagram below.)
Determine the location and size of the final image of the amoeba. The solution includes a calculation for something called the “angular magnification”—you may ignore this. I won’t cover that on your last test.
(optional) Now you switch objectives. The new objective has fo = 3 cm. The amoeba and eyepiece stay in the same place, but the new objective is now closer to the amoeba—(10 – 401/2) cm from the amoeba, and thus (14.5 + 401/2) cm from the eyepiece in order to keep the image of the new objective in exactly the same place as the image of the original objective (real microscopes have this feature). Determine the location and size of the final image, and the angular magnification in this case. (Is it bigger, smaller, or the same as before?)
EMI01. Two radio antennas A and B radiate in phase. Antenna B is 120 m to the right of antenna A. Consider point Q along the extension of the line connecting the antennas, a horizontal distance of 40 m to the right of antenna B. The frequency, and hence the wavelength, of the emitted waves can be varied.
a) What is the longest wavelength for which there will be destructive interference at point Q?
b) What is the longest wavelength for which there will be constructive interference at point Q?
EMI04. A sheet of blackened photographic film has two very narrow scratches in it that allow light to pass through. When you shine laser light of wavelength 624 nm on it, you find that the center of the first dark fringe occurs at an angle of ± 11.0o on either side of the central bright line.
a) How far apart are these scratches?
b) Find the total number of bright fringes and the total number of dark fringes that the film will produce with this laser.