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hw_ch27 - Ch 27 HW Solutions 1 Because the lens takes the...

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Ch. 27 HW Solutions 1.) Because the lens takes the parallel rays from the sun and focuses them on a point, the lens is converging. The point at which parallel rays converge is the focus of the lens, 12 cm. 3.) (a) q = 80 cm: real, inverted (b) q = 200 cm: real, inverted (c) q = -40 cm: virtual, upright 4.) (a) q = 100 cm: real, inverted (b) q = 133 cm: real, inverted (c) q = -33.3 cm: virtual, upright 8.) p = 12 cm, q = 4.0 m. f = (1 /p + 1 /q ) - 1 = (1 / 12 + 1 / 400) - 1 = 11 . 65 cm. M = - q/p = - 400 / 12 = - 33 . 3 . The image size is magnified 33.3 times, so it is 83.3 cm × 116.7 cm (inverted). 14.) q = 80 cm, p = 12 cm. f = (1 /p + 1 /q ) - 1 = (1 / 12 + 1 / 80) - 1 = 10 . 4 cm. 15.) f = 80 cm, h i = 2 m = 200 cm, h o = 50 cm. M = - h i /h o = - 200 / 50 = - 4 (real images from converging lenses are inverted). Since M = - q/p = - 4 , q = 4 p . Then 1 /f = 1 / 80 = 1 /p + 1 / 4 p = 5 / 4 p . So (a) p = (5)(80) / 4 = 100 cm. (b) q = 4 p = 400 cm. 20.) The object is 150 cm from the screen (the image location) so p + q = 150 cm. With f = 25 cm, we use the lens equation 1 25 = 1 p + 1 150 - p = 150 - p p (150 - p ) + p p (150 - p ) = 150 - p + p p (150 - p ) = 150 p (150 - p ) . 25 = p (150 - p ) 150 = - p 2 + 150 p 150 p 2 - 150 p + 3750 = 0 Using the quadratic equation, 150 ± p 150 2 - (4)(3750) 2 and solving yields p = 118 . 3 cm, 31 . 7 cm. The lens must be placed this far from the object.
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