An object is placed 10 m before a convex lens with focal length 5 . 9 m . Anothe

Question

An object is placed 10 m before a convex lens with focal length 5 . 9 m . Another convex ens is placed 9 . 39 m behind the first lens with a focal length 3 . 8 m (see the figure below). Note: Make a ray diagram sketch in order to check your numerical answer a)At what distance is the first image from the first lens? Answer in units of m
b)What is the magnification of the first image?
c)At what distance is the second image from the second lens? Answer in units of m
d)What is the magnification of the final image, when compared to the initial object?

Explanation / Answer

An object is placed 10 m before a convex lens with focal length 5.9 m. Another convex lens is placed 9.39m behind the first lens with a focal length 3.8 m.
f1 = 5.9 m
f2 = 3.8 m

Comment: We are going to apply the formulas appropriate for thin lenses.

1) The distance of the first image from the lens is computed using the formula

1/s + 1/s' = 1/f

where s is the distance of the object = 10m,

s' is the distance of the image

and f is the focal length of the first lens = 5.9m

So 1/10 + 1/s' = 1/5.9

or 1/s' = 1/5.9 - 1/10

which gives s' = 14.4 m, the distance of the first image from the first lens.

2) The magnification of the first image is computed from the formula:

M = - s'/s = -14.4/10 = -1.44.

The negative sign says that the image is inverted.

3) The object for the second lens is the image of the first lens.

This object is 14.4 - 9.39 = 5 m to the right of second lens.

In terms of the formula: s2 = -5m, and f2 = 3.8m, with the image 2 distance s2' to be determined from:

1/s2 + 1/s2' = 1/f2

or -1/5 + 1/s2' = 1/3.8

or 1/s2' = 1/3.8 + 1/5

So s2' = 2.15 m

The second image is 2.15m to the right of the second lens.

4) The magnification of the second lens only is

M2 = -s2'/s2 = -2.15/(-5) = 0.432

The combined magnification is M*M2 = -1.44*0.432 = -0.623 with the negative sign meaning

that the image is inverted.

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