Question

A) A converging lens of focal length 7.90 cm is 17.7 cm to the left of a diverging lens of focal length -5.99 cm. A coin is placed 12.5 cm to the left of the converging lens. Calculate the location of the coin's final image.

B) Calculate the magnification of the coin's final image.

Answer 1

Solution:

Let us go to the basics first.

The image of the first lens becomes the object of the second

So 1/s'₁ + 1/s₁ = 1/f₁

=>1/s'₁ = 1/f₁- 1/s₁ = 1/7.90 - 1/12.5 = 0.04658

Therefore s'₁ = 21.467cm

This means the object for the second lens is 21.467 - 17.7 = 3.767 cm behind the diverging lens

so s = -3.767 (a virtual object)

Now for the second lens we have 1/s'₂ + 1/s₂ = 1/f₂

So 1/s'₂ = 1/f ₂- 1/s₂ = 1/(-5.99) - 1/(-3.767) = 0.0985

So s'₂ = 10.15 cm (Answer)

So the image forms 10.15 cm to the right of the diverging lens

Final magnification = m1*m2 = s'₁/s₁ * s'₂/s₂ = 21.467/12.5 * 10.15/-3.767 = - 4.627 (Answer)

Thanks!!!