A diverging lens of focal length -11.5cm is placed 40.0cm from a converging lens

Question

A diverging lens of focal length -11.5cm is placed 40.0cm from a converging lens of unknown focal length. A 15.2cm tall erect object is placed 25.3cm in front of the diverging lens which is to produce an image on a screen that is twice the size of the original object but inverted.

Part A

Where should the screen be located to produce a clear image?

(Give the distance from the converging lens to the screen in cm.)

Part B

What is the focal length of the converging lens?

Part C

What is the height of the intermediate image formed by the diverging lens?

Part D

If the screen is removed and the image is viewed using your eyes instead, what angular magnification would result from placing your eyes 40 cm from the final image?

Explanation / Answer

given f1 = -11.5 cm

Apply

1/u1 + 1/v1 = 1/f1

1/v1 = 1/f1 - 1/u1

1/v1 = -1/11.5 - 1/25.3

v1 = -7.9 cm (image distance of first lense)

magnification, m1 = -v1/u1

= -(-7.9/25.3)

= 0.3125

let f2 is the focal length of second lense.

u2 = (40+7.9) = 47.9 cm (bjecte distance for second lense)

m2 = -v2/u2

over all magnification, M = m1*m2

-2 = 0.3125*(-v2/u2)

v2 = u2*2/0.3125

= 47.9*2/0.3125

= 306.6 cm

so screen distance from converging lense is 312.92 cm <<<<<-----Answer

B) Apply, 1/f2 = 1/u2 + 1/v2

1/f2 = 1/47.9 + 1/306.6

f2 = 41.42 cm <<<<<-----Answer

c) inermedite image height = m1*object height

= 0.3125*15.2

= 4.75 cm <<<<<-----Answer

D) anggular magnification, theta = tan^-1(y/V2)

= tan^-1(2*15.2/312.92)

= 5.54 degrees

= 0.097 rad

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