Physics Solution Manual for 1100 and 2101

Substituting this v expression into the second of

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: e image and object distances via the magnification equation. The image distance is give, and the object distance is unknown. However, we can obtain the object distance by using the mirror equation, which relates the image and object distances to the focal length, which is also given. SOLUTION According to the magnification equation, the magnification m is related to the image distance di and the object distance do according to m=− di do (25.4) According to the mirror equation, the image distance di and the object distance do are related to the focal length f as follows: 111 += d o di f or 1 11 =− d o f di (25.3) Substituting this expression for 1/do into Equation 25.4 gives m=− 1 1 di d = −di − = − i + 1 f d do f i 36 cm = − + 1 = −2.0 12 cm 41. SSM WWW REASONING This problem can be solved by using the mirror equation, Equation 25.3, and the magnification equation, Equation 25.4. SOLUTION a. Using the mirror equation with d i = d o and f = R / 2 , we have 1 1 1 1 1 = – = – do f di R / 2 do or 2 2 = do R Therefore, we find that d o = R . b. According to the magnification equation, the magnification is Chapter 25 Problems m=– di do =– do do = 1311 –1 c. Since the magnification m is negative, the image is inverted . 42. REASONING a. The smallest angle of incidence such that the laser beam hits only one of the mirrors is shown in the left drawing. Here the laser beam strikes the top mirror at an angle of incidence θ1 and just passes the right edge of the lower mirror without striking it. The angle θ1 can be obtained by using trigonometry. b. The smallest angle such that the laser beam hits each mirror only once is shown in the right drawing. The laser beam strikes the top mirror at an angle of incidence θ2, reflects from the top and bottom mirrors, and just passes the right edge of the upper mirror without striking it. The angle θ2 can also be obtained by using trigonometry. 17.0 cm 3.00 cm 17.0 cm θ1 3.00 cm θ2 SOLUTION a. Note from the left drawing that the length of the side opposite the angle θ1 is 1 (17.0 cm ) , 2 because the angle of reflection equals the angle of incidence. The length of the side adjacent to θ1 is 3.00 cm. Using the inverse tangent function, we find that 1 (17.0 cm ) = 70.6° 3.00 cm θ1 = tan −1 2 b. From the right drawing we see that the length of the side opposite the angle θ2 is 1 (17.0 cm ) , because the angle of reflection equals the angle of incidence. The length of the 3 side adjacent to θ2 is 3.00 cm. Again, using the inverse tangent function, we obtain θ 2 = tan −1 1 (17.0 cm ) 3 = 62.1° 3.00 cm 1312 THE REFLECTION OF LIGHT: MIRRORS 43. SSM WWW REASONING The mirror equation relates the object and image distances to the focal length. The magnification equation relates the magnification to the object and image distances. The problem neither gives nor asks for information about the image distance. Therefore, we can solve the magnification equation for the image distance and substitute the result into the mirror equation to obtain an expression relating the object distance, the magnification, and the focal length. This expression can be applied to both mirrors A and B to obtain the ratio of the focal lengths. SOLUTION The magnification equation gives the magnification as m = –di/do. Solving for di, we obtain di = –mdo. Substituting this result into the mirror equation, we obtain 1 1 1 1 1 + = + = do di do f − md o ch FI GJ HK 1 1 1 1− = do m f or or f= dom m−1 Applying this result for the focal length f to each mirror gives fA = d o mA and mA − 1 fB = d o mB mB − 1 Dividing the expression for fA by the expression for fB, we find fA fB = m c h m c − 1h 4.0b.0 − 1g 2 = = = 4 / c − 1h m c − 1h 2 .0 b.0 − 1g m m d 0 mA / mA − 1 A B d 0 mB B A B 0.67 44. REASONING The radius of curvature R of a concave mirror is related to the focal length f of the mirror by f = 1 R (Equation 25.1), so we have that 2 R=2f (1) 111 = + (Equation 25.3), f d o di where do = 14 cm is the object distance and di is the image distance. The image distance is d related to the object distance via the magnification equation m = − i (Equation 25.4). do Because the image is virtual, the image distance di is negative. The object distance is positive, so we conclude from Equation 25.4 that the magnification is positive: m > 0. The image is twice the size of the object, so we have that the magnification is m = +2.0. The mirror’s focal length is given by the mirror equation Chapter 25 Problems SOLUTION Taking the reciprocal of both sides of f= 1 11 + d o di 1313 111 = + (Equation 25.3) yields f d o di . Substituting this result into Equation (1), we find that R=2f = Solving m = − 2 (2) 1 1 + d o di di (Equation 25.4) for di and substituting m = +2.0 yields do di = − md o = −(+2.0)d o = −2d o (3) Substituting Equation (3) into Equation (2), we obtain R= 2 1 1 + do di = 2 1 1 + d o ( −2 d o ) = 2 = 4d o = 4 (14 cm ) = 56 cm 1 2d o 45. REASONING Since the size hi of the image is one-fourth the size ho of the object, we know from...
View Full Document

This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

Ask a homework question - tutors are online