Lenses/Images - Find final image

[maniacp08]

Lenses/Images -- Find final image

Two converging lenses, each having a focal length equal to 8 cm, are separated by 35 cm. An object is 30 cm to the left of the first lens.

(a) Find the position of the final image using both a ray diagram and the thin-lens equation

(c) What is the overall lateral magnification?

What I did:
s1' = f1s1/(s1-f1)
s1' = (8cm)(35cm)/(35cm-8cm) = 10.37cm

m1 = - s1'/s = -10.37cm/35cm = -.296285cm

s2' = f2s2/(s2-f2)
s2 = 35cm-30cm = 5cm
s2' = (8cm)(5cm)/(5cm-8cm) = -13.333cm

m2 = - s2'/s = - (-13.333cm)/5cm = 2.666cm

For question a isn't it 35cm+s1'+s2'=32.04cm or 32cm? This answer was marked wrong can anyone tell me what I did wrong?

For question c - m = m1*m2 = -.296285cm*2.666cm = -.78989581cm but this is again wrong

Please help, thanks!

 

[Carid]

The first image is formed at 10.37cm to the right of the first lens.
This image is now the object for the second lens.
How far is it from the second lens?

 

[nlightner]

Based on the given information, the final image is formed at a distance of 32 cm to the right of the second lens. This can be determined using the thin-lens equation: 1/f = 1/s1 + 1/s2, where f is the focal length of the lens, and s1 and s2 are the object and image distances, respectively. Plugging in the given values, we get 1/8 = 1/30 + 1/s2, which gives s2 = 40 cm. Since the object is 30 cm to the left of the first lens, the final image is formed at a distance of 30 cm + 40 cm = 70 cm to the right of the first lens. This can also be confirmed using a ray diagram, where the rays from the object will converge at this point.

For the lateral magnification, we need to take into account the magnification of both lenses. The overall lateral magnification is given by the product of the individual magnifications, which in this case is (-0.296285)(2.666) = -0.7898. This indicates that the final image is smaller and inverted compared to the original object. However, the negative sign indicates that the final image is on the opposite side of the lens, which is to the right in this case. So the final image is smaller and inverted, located 70 cm to the right of the first lens.