Question

An astrophysicist mounts two thin lenses along a single optical axis (the lenses are at right angles to the line connecting them, and they appear concentric when viewed from either end). The lenses are identical, each with a positive (converging) focal length of 15.2 cm. They are separated by a distance of 40.2 cm. Lens 1 is to the left of Lens 2. In order to evaluate the lens combination as a single optical instrument, the teacher places an object 30.0 cm in front of (to the left of) Lens 1.

Required:
a. What is the final image's distance (in cm) from Lens 2? (Omit any sign that may result from your calculation.)
b. Where is the final image located?
c. What is the overall magnification of the lens pair, considered as a single optical instrument?

Answer 1

Answer:

1 / i + 1 / o = 1 / f     thin lens equations

i = o f / (o - f)   rearranging

Lens 1:   object = 30 cm    f = 15.2 cm

i1 = 30 * 15.2 / (30 - 15.2) = 30.8 cm

o2 = 40.2 - 30/8 = 9.4 cm    distance of image 1 from lens 2

i2 = 9.4 * 15.2 / (9.4 - 15.2) = - 24.6 cm

The final image is 24.6 cm to the left of lens 2

The first image is inverted

The second image is erect (as seen from the first image)

So the final image is inverted

M = m1 * m2 = (-30.8 / 30) * (24.6 / 9.4) = -2.69

Answer 2

It can be deduced that the final image's distance from Lens 2 will be 30.8 cm.

How to calculate the distance

By using the Lens formula, the distance will be calculated thus:

1/v + 1/u = 1/15.2

1/v + 1/30.0 = 1/15.2

v = 30.8cm

In this case, the image formed will be to the right.

Lastly, the overall magnification of the lens pair will be:

M = (30.8/30.0)[(-28.4/40.7 - 30.8)]

M = -2.94.

In conclusion, the magnification is -2.94.

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