Question

A converging lens with a focal length of +30 cm is placed 40 cm to the left of a diverging lens with a focal length of -40 cm. An object is placed 60 cm to the left of the converging lens. What is the location and orientation of the image produced by the combination of lenses?

Why is the answer B?

(a) Upright, on the right of the diverging lens and larger than the main object

(b) Inverted, on the right of the diverging lens and larger than the main object

(c) Inverted, on the right of the diverging lens and smaller than the main object

(d) Upright, on the left of the diverging lens and larger than the main object

(e) Inverted, on the left of the diverging lens and smaller than the main object

Answer 1

Using the lens equation for first converging lens

1/f = 1/u + 1/v

u = object distance = +60 cm

f = focal length = +30 cm

v = image distance = ?

1/v = 1/30 - 1/60

v = 30*60/(60 - 30) = 60 cm

Now this image will be +60 cm right from the converging lens, Image's distacne from diverging lens will be

u1 = 40 - 60 = -20 cm = object distnace for diverging lens

f1 = focal length of diverging lens = -40 cm

v1 = image distance = ?

1/v1 = -1/40 + 1/20

v1 = 40*20/(40 - 20) = 40 cm

v1 = 40 cm (+ve sign means image to the right of diverging lens and will be inverted)

Part B.

Magnification is given by:

M = M1*M2 = (-v/u)*(-v1/u1)

M = (-60/60)*(-40/(-20))

M = -2, Since M > 1, So image will be larger than main object.

Correct option is B.

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