Question on a diverging lens?? Thanx

Question

If a diverging lens of focal length -61.0 cm is then placed 28.0 cm from the first lens on the opposite side of the lens from the object, what is the location of the new image? What is the magnification due to the two-lens system?

Answer 1

I assume this question follows on from your previous one, as it is not complete in itself.

The real image from the convex lens was:

59.4 - 28.0

= 31.4 cm.

beyond the diverging lens, which now stops it from forming in that position.

As the light is converging towards this lens, it now gives a virtual object at distance 31.4 cm. from this lens. If v is the new image distance from the second lens:

- 1 / 31.4 + 1 / v = - 1 / 61.0

v = 64.7 cm.

The final image is real, and 64.7 cm. from the diverging lens on the opposite side from the converging lens.

The magnification m of the two lens system is the product of the magnifications from each lens:

m = (59.40 / 34.94) * (64.72 / 31.40)

= 3.50.