Homework 13 chapter 36: 17, 21, 33, 74
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1 Homework chapter 6: 7,,, 74
2 Problem 6.7 A spherical mirror is to be used to form, on a screen located 5 m from the object, an image five times the size of an object. (a) Describe the type of mirror required. (b) Where should the mirror be positioned relative to the object? One can solve the problem formally from the mirror equation (if we remember the equation!). It is however, a good idea to imagine the situation graphically, before attempting the formal solution. The image on a screen can only be formed in front of the mirror. The image must be real. It is suggested in the problem that the object is also real. We can draw a ray diagram (backwards) to illustrate the situation. I F d 5 m O s
3 Since we know the location of the screen, we should start our analysis by drawing the image. The only ray we can track at this time is the chief ray. We know that the object must be somewhere along the object's chief ray. We have to look for a point on the object's chief ray, which is five times closer to the principle axis than the image is. From the similarity of the triangles formed by the principle axis, the chief rays, the image and the object, we can immediately figure out that the object must be five times closer to the mirror than the image. After we located the position of the object, we can track both parallel rays, and graphically locate the focal point. Let's now find the formal solution. In order to have a real image enlarged five times, the magnification of the mirror in this situation must be -5. We can either recall the relationship between the magnification of a mirror (difficult to remember) or consider the two similar triangles in the ray diagram formed by the principal axis, the chief rays, the object and the image. s s d () m ' + s s b) This equation contains the only unknown requested in this part of the problem. Solving for the position of the object we find: d m s m m + 5 ( 5) + 5. a) To describe the mirror we can use any characteristic of the mirror. Let's for example find the focal length of the mirror. As taken directly from the mirror equation
4 () + s s ' f the focal length of the mirror can be expressed as below ( s + d) ( s + d) s s.5m 6.5m f.04m s + s +.5m + 6.5m We can also obtain this result from geometrical considerations. From the similarity of the triangles formed by the principle axis, the object focal ray, the object and the mirror s f h ( ) f h ' m Therefore ms m f m ( m ) ( 5 ) Positive focal length indicates a concave mirror. (The curvature radius of the mirror is: R f. 08m) 4
5 Problem 6. A cubical block of ice 50 cm on a side is placed on a level floor over a speck of dust. Find the location of the image of the speck if the index of refraction of ice is.09. β n A B I t n dβ t α dα γ O This is a case where an image is not really formed. The refracted rays and their extensions do not meet at a single point. We can see the speck of dust only because we observe the light in a very small (differential) cone determined by the size of our pupils. Only a small (differential) fragment (AB) of the surface forms the observed image. (You have probably noticed in nature, that in such an arrangement, the depth at which the image is observed depends on the observation angle β.) Although the text suggests solving the problem for a direct from above observation, the solution presented here considers a general case. In order to find the depth at which the image I is observed one has to relate the dimensions of the (differential) triangle ABI with the dimensions of the (differential) triangle ABO. Both triangles have one angle which is very small (differential size). From the sine theorem 5
6 AB OA sin dα sin 80 ( γ) Geometrical considerations enable one to express the distance OA in terms of the object s depth t t OA cos α and angle γ with the angle of incidence γ ( 90 + α + dα) Expanding the trigonometric in Maclaurin series, one finds the first significant terms to be sin dα sin 0 + cos0 dα dα sin ( 80 γ) sin( 90 ( α + dα) ) cos( α + dα) cosα Solving the equations for AB t AB dα cos α Identical analysis of triangle ABI relates dimension AB with the depth t of the image t' AB dβ cos β Combining the last two equations eliminates AB and relates the depths cos β dα t' t cos α dβ Using Snell s law one can eliminate the dependence of the image depth on one of the angles. Differentiation of Snell s law n sin α n sin β, 6
7 relates the differentials of the angles n cosα dα n cosβ dβ and a direct transformation the angles themselves Therefore cosα cos t' t cos sin β n α n n α n cos t n n Note that the apparent depth t of the object has the highest value when looked at directly from above (β0 ). In this case t' 50cm cos.09 0 sin 0 β.09 sin sin β β 50cm.09 n n 8.cm As the observation angle β increases, the image appears at a shallower and shallower depth. 7
8 Problem 6. The nickel s image in Figure P6. has twice the diameter of the nickel and is.84 cm from the lens. Determine the focal length of the lens. I Though a magnifying glass one looks at a virtual image. A O negative (opposite to the distance from the lens) number F i must therefore describe the F o position of the image. Considering the chief ray, one should recognize that in order s to have the magnification of m, the object must be placed s -.84 cm exactly at the midpoint between the lens and the image. Therefore the object is at position The lens equation gives s m f + s m + m m.84cm.84cm Note that an accurate ray diagram indicates that the object focal point is at the location of the image. The position of the object focal point at the same location of the image is coincidental and results from this particular magnification. 8
9 Problem 6.74 Two converging lenses with focal lengths of 0 cm and 0 cm are positioned 50 cm apart, as shown in Figure P6.74. The final image is to be located between the lenses, at the position indicated. (a) How far to the left of the first lens should the object be? (b) What is the overall magnification? (c) Is the final image upright or inverted? object h F i final image F o h' first image H p q l cm s d 50 cm A ray diagram helps us to imagine the situation. Since the final image (formed by the second lens) is not on the same side as the image rays (refracted be the second lens) the lens forms a virtual image. Tracing the chief ray and the image parallel ray (and the image focal ray) allows one to determine the position of the object for the second lens (image formed by the first lens). The found object (first image) is on the same side as the object rays therefore the object is real. Tracing the image chief, focal (and parallel) rays of the first lens allows one to identify the initial object. The algebraic solution can be found from the lens equation applied to both lenses. Consistent with the figure ) + p q f 9
10 ) + s s ' Additionally ) q+ s d 4) + l d f We obtained four equations with four unknown. The rest is algebra. We have to solve the above set for p. From the information given (eq. 4) we can determine the position of the final image (with respect to the second lens) d l 50cm cm 9 cm ( ) ( ) We can use this information in equation and find the position (with respect to the second lens) of the first image s s f 0cm -9cm ' 974. cm The position of the first image with respect to the first lens is therefore (from eq. ) q d s 50cm cm Now from equation p q f 0cm 40.6cm. cm (It makes sense to perform calculation step-by-step. We can check each step for possible discrepancies with the ray diagram.) 0
11 b) Considering similar triangles, formed by the optical axis, the chief rays, the objects and the images, one finds the following two proportionalities h' H s and H h q p From its definition, the magnification of the system is M h' h h' H H h s q p 9.0cm 9.74cm 40.cm.cm 5.9 c) Negative magnification, as well as the ray diagram, indicates that the final image is inverted.
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