Two tenses that are 1.00 m apart are used to form an image, as shown. Lens 1 (on
ID: 1377482 • Letter: T
Question
Two tenses that are 1.00 m apart are used to form an image, as shown. Lens 1 (on the Left) is converging and has a focal length of 15.0 cm. Lens 2 (on the right) is diverging and has a focal length of -70.0 cm. The object is placed 30.0 cm to the left of lens 1. The two tenses have the same optical axis, Find (a) the location of the first image (due to Lens 1), (b) the location of the final image (due to lens 2), (c) The magnification of the final image relative to the (initial) object. (d) Is the final image real or virtual? Is it upright or inverted with respect to the original object?Explanation / Answer
Solution:
a) Focal length of converging lens = fc = 15 cm
object distance for the convex lens = pc = 15 cm
image distance for the convex lens = qc
1/pc +1/qc = 1/fc
1/qc = 1/fc - 1/pc
= 1/15 - 1/30
=1/30
=> qc = 30 cm image distance for the convex lens. (real image for lens 1 and is of same size as the object)
This image becomes object for lens 2 .
b) Focal length of concave lens (lens 2) = fd = -70cm
Object distance for the concave lens pd= 1m - 0.30m = 0.70 m =70 cm
since a diverging lens forms a virtual image, the image distance is negative
1/pd + 1/qd = 1/fd
=> 1/qd =1/fd - 1/ pd
= 1/- 70 - 1/70
= -2/70
= -1/35
=>qd = -35 cm = image distance for the lens 2.
c) Magnification = - qd/qc = -(-35)/ (100-35) = 0.54
=> image size is 0.54 times the origiinal object.
d) It is inverted with respect to original object and is also diminished.
The final image is virtual, since it is formed by the concave lens.
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