PHY_solution-chapter34

PHY_solution-chapter34 - 34.1: If up is the y +-direction...

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Unformatted text preview: 34.1: If up is the y +-direction and right is the x +- direction, then the object is at ), , ( at is ), , ( 2 y x P y x- ′-- and mirror 1 flips the y-values, so the image is at ) , ( y x which is . 3 P ′ 34.2: Using similar triangles, m. 3.24 m 0.350 m 0.350 m 28.0 m 040 . mirror tree mirror tree mirror tree mirror tree = + = = ⇒ = d d h h d d h h 34.3: A plane mirror does not change the height of the object in the image, nor does the distance from the mirror change. So, the image is cm 2 . 39 to the right of the mirror, and its height is cm. 85 . 4 34.4: a) cm. . 17 2 cm . 34 2 = = = R f If the b) spherical mirror is immersed in water, its focal length is unchanged—it just depends upon the physical geometry of the mirror. 34.5: a) b) cm, . 33 cm 5 . 16 1 cm . 22 2 1 1 1 1 = ′ ⇒- = ′ ⇒ = ′ + s s f s s to the left of the mirror. cm, 20 . 1 cm 16.5 cm 33.0 cm) 600 . (- =- = ′- = ′ s s y y and the image is inverted and real. 34.6: a) acb91f1c69e9a12614ac6063bd5ec3610bcb8bb2.doc Page 1 of 37 b) cm, 60 . 6 cm 5 . 16 1 cm . 22 2 1 1 1 1- = ′ ⇒-- = ′ ⇒ = ′ + s s f s s to the right of the mirror. cm, 240 . cm 16.5 cm) .60 6 ( cm) 600 . ( =-- = ′- = ′ s s y y and the image is upright and virtual. 34.7: m. 75 . 1 m 10 58 . 5 1 m 75 . 1 1 1 1 1 1 10- = ′ ⇒ ×- = ′ ⇒ = ′ + s s f s s m. 10 13 . 2 ) m 10 6794 )( 10 14 . 3 ( 10 14 . 3 10 58 . 5 75 . 1 4 3 11 11 10--- × = × × = = ′ ⇒ × = ×- = ⇒ my y m 34.8: cm 40 . 1 cm . 21 1 cm 00 . 3 2 1 1 1 1 , cm 00 . 3- = ′ ⇒-- = ′ ⇒ = ′ +- = s s f s s R (in the ball). The magnification is . 0667 . cm . 21 cm 40 . 1 =-- = ′- = s s m 34.9: a) . Also . 1 1 1 1 1 1 s f f s s m f s sf s fs f s s f s f s s- = ′- =- = ′ ⇒- =- = ′ ⇒ = ′ + For b) , , ′ ⇒ s f s f so the image is always on the outgoing side and is real. The magnification is , <- = s f f m since . s f < For c) , 1 2 =- < ⇒ ≥ f f m f s which means the image is always smaller and inverted since the magnification is negative. For . 1 2 = ⇒ <- < ⇒ < < f f m f f s f s f Concave d) mirror: , < ′ ⇒ < < s f s and we have a virtual image to the right of the mirror. , 1 = f f m so the image is upright and larger than the object. acb91f1c69e9a12614ac6063bd5ec3610bcb8bb2.doc Page 2 of 37 34.10: For a convex mirror, . < +- =- = ′ ⇒ < f s f s f s sf s f Therefore the image is always virtual. Also , + =--- =- = s f f s f f s f f m so the image is erect, and , since 1 f s f m + < so the image is smaller. acb91f1c69e9a12614ac6063bd5ec3610bcb8bb2.doc Page 3 of 37 34.11: a) b) . , for < ′ s f s s c) . for f s s < < < ′ If the object d) is just outside the focal point, then the image position approaches positive infinity....
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This note was uploaded on 02/25/2009 for the course PHY 127 taught by Professor Mr.yang during the Spring '08 term at SUNY Stony Brook.

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PHY_solution-chapter34 - 34.1: If up is the y +-direction...

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