35_Ch 23 College Physics ProblemCH23 Mirrors and Lenses

# 35_Ch 23 College Physics ProblemCH23 Mirrors and Lenses -...

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Mirrors and Lenses 301 (c) When 1.5 n = and , 1 0 2 2.00 n = ( ) ( ) 6.00 m 2.00 24.0 m 2.00 1.50 f = = + and ( )( ) 10.0 m 10.0 m 24.0 m 17.1 m 24.0 m p f p f = = =− q The image is 17.1 m t of the lens to the lef (d) Observe from equation (1) that 0 f < if n and when n . Thus, a diverging lens can be changed to converging by surrounding it with a medium whose index of refraction exceeds that of the lens material. 1 n > 2 2 0 f > 1 n < 23.62 The inverted image is formed by light that leaves the object and goes directly through the lens , never having reflected from the mirror. For the formation of this inverted image, we have 1 1 1.50 q M p = − = − giving 1 1 1.50 q p = + The thin lens equation then gives 1 1 1 1 1 1.50 10.0 cm p p + = or ( ) 1 10.0 cm 16.7 cm 1 1 16.7 cm 1.50 = 23.3 cm = + p = + The upright image is formed by light that passes through the lens after reflecting from the mirror. The object for the lens in this upright image formation is the image formed
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