A converging lens of focal length f1 = +22 .5cm is placed at a distance d = 60 .0cm to the left of a diverging lens of focal length f2 = −30.0cm. An object is placed on the common optical axis of the two lenses with its base 45.0cm to the left of the converging lens.
(a) Calculate the location of the final image and its overall magnification with respect to the object.
Lens A
1/u1 + 1/v1 = 1/FA
1/45.0cm + 1/v1 = 1/22.5cm
v1 = 45.0cm
This value is positive so the image is real.
Lens B
u2 = d−v1 = 60.0cm−45.0cm = 15.0cm
Again this value is positive so the image is real.
1/u2 + 1/v2 = 1/FB
1/15.0cm + 1/v2 = 1/30.0cm
v2 = −30.0cm
This value is negative so it is a virtual image.
Magnification
m = v1/u1 × v2/ u2 m = 45.0cm/45.0cm × 30.0cm/15.0cm
m = 2×
1 answer
The final image is located 30.0cm to the right of the diverging lens and has an overall magnification of 2×.
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