1. An object is 125.0 cm from a screen. A converging lens of focal length 15.0 c

Question

1. An object is 125.0 cm from a screen. A converging lens of focal length 15.0 cm is to be placed between the object and the screen so that the largest possible image is clearly focused on the screen. Where should the lens be placed, and what magnification results?
2. A pair of converging lenses, each of focal length 15.0 cm, are placed 25.0 cm apart. An object of height 2.0 cm is placed 20.0 cm from the first lens.
(a) Locate the final image, and determine its height.
(b) Confirm your calculations by constructing a ray trace.

Explanation / Answer

let,
s = d cm
s' = (125-d) cm
f = 15 cm

use:
1/f = 1/s' + 1/s
1/15 = 1/(125-d) + 1/d
1/15 = 125/((125-d)*d)
125*d - d^2 = 125*15
d^2 -125*d + 1875 = 0
solving above quadratic equation,
d = 107.6 cm and x =17.4 cm

for largest magnification, object should be placed between f and 2f
so,
lens should be placed at a distance of 17.4 cm from object

m = -s'/s
= (125-d)/d
= (125-17.4)/17.4
= 6.18
magnification is 6.18

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