Two convex lenses A and B of focal lengths 20 cm and 10 cm are placed coaxially 10 cm apart. An object is placed on the common axis at a distance of 10 cm from lens A. Find the position and magnification of the final image. - Zigya
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Two convex lenses A and B of focal lengths 20 cm and 10 cm are placed coaxially 10 cm apart. An object is placed on the common axis at a distance of 10 cm from lens A. Find the position and magnification of the final image.


From figure below, we have, for lens A
f1 = + 20 cm and u1 = –10 cm
The image distance v1 is given by
which gives                      1 over straight v subscript 1 space equals space 1 over straight f subscript 1 plus 1 over straight u subscript 1 space equals space 1 over 20 plus fraction numerator 1 over denominator left parenthesis negative 10 right parenthesis end fraction space equals 1 over 20 minus 1 over 10
                                            straight v subscript 1 space equals space minus 20 space cm
Thus, a virtual image is formed at I1 at a distance of 20 cm from lens A, if the lens B were absent. This image acts as a virtual object for lens B which forms the final image at I2 at a distance v2 from lens B.
For lens B we have,
x = 10 cm, u2 = – (20 + 10) = –30 cm f2 = + 10 cm.

From figure below, we have, for lens Af1 = + 20 cm and u1 = –10 c

The image distance v2 is given by
                               1 over straight v subscript 2 space equals space 1 over straight f subscript 2 plus 1 over straight u subscript 2 space equals space 1 over 10 minus 1 over 30 space equals space 1 over 15
which gives                 straight v subscript 2 space equals space plus space 15 space cm
Thus, a real image I2 is formed at a distance of 15 cm from lens B.
Magnification due to straight A left parenthesis straight m subscript 1 right parenthesis space equals space straight v subscript 1 over straight u subscript 1 space equals space fraction numerator negative 20 over denominator negative 10 end fraction space equals space plus 2
Magnification due to straight B left parenthesis straight m subscript 2 right parenthesis space equals space straight v subscript 2 over straight u subscript 2 space equals fraction numerator 15 over denominator negative 30 end fraction space equals space minus 1 half
Magnification of the final image is
                                  straight m space equals space straight m subscript 1 space cross times space straight m subscript 2 space equals space plus 2 space cross times space left parenthesis negative 1 divided by 2 right parenthesis space equals space minus 1
This shows that the final image is inverted and is of the same size as the object.

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