Ch. 23 Homework, pt.4 Begin Date: 330/2018 20000 PM-Due Date: 4/2/2018 5:0000 PM

Question

Ch. 23 Homework, pt.4 Begin Date: 330/2018 20000 PM-Due Date: 4/2/2018 5:0000 PM End Date: 44/2018 9 (20%) Problem 4: An object of height 2.7 cm is placed 28 cm in front of a diverging lens or focal length 13 cm. Behind the diverging lens, and 13 m from it, there is a converging lens of the same focal length signment Status Click here for detailed view Find the location of the final image, in centimeters beyond the converging lens. ? S0% Part (a) Grade Summary s" roblem StAtus sino cos7 8 9 1 Completed 2 Completed Atets remainine g per awemp atano acotan sinh cosh) Degrees Radians SubitHlint sve upl Hints b deducton per hHi remaining 2 Feodback t doduction per foodbac 50% Part (b) what is the magnification of the final image? Include its sign to indicate its orientation with respect to the object

Explanation / Answer

Object distance = u = 28 cm

focal length = f = - 15 cm

use the lens formula

1/f = 1/v + 1/u

=> - 1/15 = 1/v + 1/28

=> v = - 9.767 cm

this is the image distance from the diverging lens which is on the same side as the object.

Object distance to the converging lens now will be:

u' = 13 - 9.767 = 3.232 cm

f' = + 15 cm

so,

1/15 = 1/v' + 1/3.232

=> v' = s'' = - 4.12 cm

because the object was on the right, and the image distance is found to have a negative sign, the final location of image will be to the right of the converging lens.

this is the location of the image beyond the converging lens.

b] Magnification of the final image will be:

M = (-v/u)(-v'/u') = (-(-9.767)/28) (-(-4.12)/3.232) = + 0.1446.

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