1. 418 points Previous Answers SerPSE8 36.P.078 My Notes An object 1.99 cm high

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1. 418 points Previous Answers SerPSE8 36.P.078 My Notes An object 1.99 cm high is placed 40.2 cm to the left of a converging lens having a focal length of 30.5 cm. A diverging ens with a foca ength of -20.0 cm is placed 11 0 cm to the right of the converging lens (a) Determine the position of the final image. 16.23 distance Your response differs from the correct answer by more than 10%. Double check your calculations. cm location to the right of the diverging lens (b) Determine the magnification of the final image. (c) Is the image upright or inverted? inverted O upright (d) Repeat parts (a) through (c) for the case in which the second lens is a converging lens having a foca length of 20.0 cm. final image location 0.96 distance Your response differs from the correct answer by more than 10%. Double check your calculations. cm location to the right of the second lens magnification Is the image upright or inverted? inverted upright Need Help? Read it iChat About It

Explanation / Answer

a) Using lens for converging lens,

1/f = 1/i + 1/o

1/30.5 = 1/i + 1/40.2

i = 126.40 cm

for 2nd diverging lens,

image from 1st lens will work as object for 2nd lens

distance of object for 2nd lens = 126.40 - 110 = 16.40 cm

1/-20 = 1/16.40 + 1/i

i = 9.01 cm right from 2nd lens

and 9.01 + 110 = 119.01 cm to the rigt from first lens

and 119.01 + 40.2 = 159.21 cm to the right from object

b)
magnification = M1*M2

= (-126.40/40.2) (9.01 /16.40)

= - 1.73

c) inverted.

d) 1/20 = 1/16.40 + 1/i

i = 91.11 cm right from 2nd lens

and 91.11 + 110 = 201.11 cm to the rigt from first lens

and 201.11 + 40.2 = 241.311 cm to the right from object

magnification = (-126.40/40.2) (91.11 /16.40)

= -17.46

inverted

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