Mirrors and Lenses301(c) When 1.5n=and , 1022.00n=( )( )6.00 m2.0024.0 m2.001.50f==+−and ()( )10.0 m10.0 m−−24.0 m17.1 m24.0 mpf=−qThe image is 17.1 mt of the lensto the lef(d) Observe from equation (1) that 0f<if nand 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. 1n>220f>1n<23.62The 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 111.50qMp=−giving 111.50qp= +The thin lens equation then gives 11.5010.0 cmpp+=or 110.0 cm16.7 cm−1116.7 cm1.50=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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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.