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.

Solution

Given, two convex lenses A and B.
Focal length of first lens, f_A = + 20 cm
Object distance from lens A, u₁ = –10 cm
The image distance v₁ is given by,

\frac{1}{v_{1}} = \frac{1}{f_{1}} + \frac{1}{u_{1}} = \frac{1}{20} + \frac{1}{(- 10)} = \frac{1}{20} - \frac{1}{10}

v_{1} = - 20 cm

Thus, a virtual image is formed at I₁ at a distance of 20 cm from lens A, in case, if the lens B were absent.

Now, 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,
Object distance from second lens, u₂ = – (20+10) = –30 cm
Focal lengtn of lens B, f_B = +10 cm.

The image distance v₂ is given by

\frac{1}{v_{2}} = \frac{1}{f_{2}} + \frac{1}{u_{2}} = \frac{1}{10} - \frac{1}{30} = \frac{1}{15}

i.e.,

v_{2} = + 15 cm

Thus, a real image I₂ is formed at a distance of 15 cm from lens B.

Magnification due to lens A, m₁

= \frac{v_{1}}{u_{1}} = \frac{- 20}{- 10} = + 2

Magnification due to lens B, m₂

= \frac{v_{2}}{u_{2}} = \frac{15}{- 30} = - \frac{1}{2}

∴

Magnification of the final image is

m = m_{1} \times m_{2} = + 2 \times (- 1 / 2) = - 1

Since the magnification is negative, the final image is inverted and is of the same size as that of the object.

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Wave Optics

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.

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