HOMEWORK 4 with Solutions


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1 Winter 996 HOMEWORK 4 with Solutions. ind the image of the object for the single concave mirror system shown in ig. (see next pages for worksheets) by: (a) measuring the radius R and calculating the focal length for the concave mirror, (b) drawing the focal point into the diagram, and (c) constructing the image of the object graphically with three rays. R = 0 cm, and f = R 2 = 5 cm Three rays: Incident Rays Reflected Rays Parallel to the optical axis Through the focal point of the mirror () 2 Through the center of the mirror (C) Through the center of the mirror (C) 3 Through the focal point of the mirror () Parallel to the optical axis C Object
2 2. Regarding your own face as a real object, describe the image of your face which you see standing 4 feet from the center (and looking directly toward) a polished and wellreflecting brass ball 2 feet in diameter hanging in front of a pawn shop. Determine the image of your face both graphically and by calculation. By Calculation: Brass ball is equivalent to a CONVEX MIRROR ( f < 0). Diameter = 2 R = 2 ft R = ft and f = 0.5 ft You are standing at 4 ft from the center of the brass ball, but the distance s o of object from the mirror is measured from the vortex of mirror. s o = (4 ) = + 3 ft rom the mirror equation, f = s o + s i = { }  = 0.43 ft Since s i < 0, the image is virtural. y i M = y o = s i s o = Since M > 0 and M <, the image is upright and minified. s i Graphically, ft s o 4 ft C
3 3. (A) Calculate the focal length of a thin biconcave lens (n=.6) in air, having radii of 6 and 8 cm. (B) Locate and describe (magnified? inverted? real?) the image of a real object 5 cm from lens. (C) What changes if the object is virtual instead of real? (D) What changes in (A), (B), and (C) if everything is immersed in water (n=.33)? (A) biconcave lens R < 0 and R 2 > 0, negative lens. R = 6 cm, R 2 = +8 cm, n =.6 By the lens maker's equation, f = (n ) { R R 2 } f = 7.5 cm (B) s o = + 5 cm. By the lens equation, f = s o + s i s i = 3.0 cm y i s i M = y o = s o = is at 3.0 cm in the left side of the lens, virtual ( s i < 0 ), upright ( M > 0 ), minified ( M < ). (C) or a virtual object, the sign of s o is changed from + to. s o = 5 cm. Then, s i = cm. M = 3.0 is at 5.0 cm in the right side of the lens, real ( s i > 0 ), upright ( M > 0 ), magnified ( M > ). (D) In water (n o =.33), the lens maker's equation is changed to n o f = (n n o ) { R R 2 } f = 22.2 cm or a real object at 5 cm, s i = 4. cm and M = is at 4. cm in the left side of the lens, virtual ( s i < 0 ), upright ( M > 0 ), minified ( M < ). or a virtual object at 5 cm, s i = cm and M = +.3. is at 6.5 cm in the right side of the lens, real ( s i > 0 ), upright ( M > 0 ), magnified ( M > ).
4 4. A positive meniscus thin lens (n=.48), with radii of curvature 4 and 2 cm, is positioned in contact with a planoconcave lens (n=.4) of radius 0 cm. (A) What is the effective focal length and refractive power of the lens combination? (B) What image (real? magnified? inverted?) will this lens combination produce from a real object located 5 cm away from it? (C) What will be the focal length of the lenscombination if the two lenses are separated by D = 5 cm? Positive meniscus thin lens: R = + 4 cm and R 2 = +2 cm (or R = 2cm, R 2 = 4cm) f = (n ) { R R 2 } f = cm Planoconcave lens: R = and R 2 = +0 cm (or R = 0 cm and R 2 = ) f 2 = 25 cm (A) f eff = f + f 2 f eff = + 25 cm Effective refractive power = Inverse of effective focal length (in meter) = 4 m  = 4 Diopters (B) s o = + 5 cm s i = 6.25 cm, and M = s i / s o = +.25 is located at 6.25 cm in the left side of the lens combination, virtual (s i < 0), upright ( M > 0), and magnified ( M > ).
5 (C) D = 5 cm f ( D f 2 ) f f = D (f + f 2 ) = cm (in the left side of the first lens) f b = f 2 ( D f ) D (f + f 2 ) = 2.27 cm (in the left side of the second lens) *** Effective focal length D f eff = f + f 2 f f 2 f eff = +.4 cm Then, the principal planes are at L H = 6.8 cm and L 2 H 2 = 3.6 cm 0 cm L L 2 f * * b 2 2 H H 2 feff feff
6 5. Two thin lenses (L doubleconvex with focal points, and L 2 doubleconcave with focal points 2 ) are placed at a distance D as shown in ig.2 (see next pages for worksheets). Determine the image of object S using a "twostep process" as discussed in class. Obtain location and magnification of image (A) Graphically, using the figure below (B) By numerical calculation using thin lens equations for L and L 2 If done right (with proper signs) results from (A) and (B) should agree. You may compare your results with the one obtained using the equation for s i2 derived and formulated on page 5 of Handout I. (Use the appended worksheet to find the graphical solution.) (A) Graphically, L L 2 of L2 S of L Object of L2 2 2 (B) By calculation, (All parameters are measured by a ruler.) f = cm, f 2 = 2.5 cm, D = 6.0 cm, s o = cm, y o = cm or the lens L, s i = cm and y i = 3 cm. of L is real, inverted, and magnified. or the lens L 2, s o2 = D s i = 6 cm and y o2 = y i = 3 cm. s i2 = 4.3 cm and y i2 = cm of L 2 is virtual, upright, and magnified (compared to the S). Good agreement with Graphical solution! ** rom the page 5 of Handout I, D s i2 = f 2 D (s o f ) f f 2 s o (D f 2 )(s o f ) s o f = 4.3 cm (Good agreement!)
7 (A) Air Air Object C C 2 n =.65 (B) Water n=.33 Water C C 2 (C) n=.7 Air Air C Virtual Object n=.4
8 7 *. Optional ind graphically the image of the object for two plane mirror system shown in ig.4 (see next pages for worksheets) where two mirrors are inclined to each other at 90. Draw all possible images and two rays (at least) for each image. Object 90
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