An object 2.00 cm high is placed 61.1 cm to the left of a converging lens having

Question

An object 2.00 cm high is placed 61.1 cm to the left of a converging lens having a focal length of 56.1 cm. A diverging lens having a focal length of ?20.0 cm is placed 110 cm to the right of the converging lens. (Use the correct sign conventions for the following answers.)

(a) Determine the final position and magnification of the final image. (Give the final position as the image distance from the second lens.)

final position _____cm

Select:

to the left of first lens

between the two lenses

to the right of the second lens

magnification ________

(c) Repeat parts (a) and (b) for the case where the second lens is a converging lens having a focal length of +20.0 cm.

Position ______cm

Select:

to the left of first lens

between the two lenses

to the right of the second lens

magnification ________

THANK YOU

Explanation / Answer

we know that

Form converging lense

1/f = 1/p +1/q

1/q = 1/f - 1/p

q = pf /( p-f) = 61.1(56.1) / (61.1-56.1) = 685.5cm

This image will act as the object for the diverging lense

-1/f = 1/p+1/q2

1/q2 = -1/f-1/p

In this case p=q-110cm = 685.5-110 = 575.5cm

1/q2 = -1/20 - 1/575

q2 = -19.32cm ( This will be the final image. This will be formed to the right of second lense)

Magnification = Image size / object size

= -19.32cm / 2cm = -9.6cm

Same procedure will be adapted for the second question.

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