Question

A symmetric converging lens is used in an experiment and the observations are tabulated.
Object distance Image distance
10 cm 30 cm
20 cm 60 cm
30 cm 30 cm
60 cm 20 cm
45 cm 22.5 cm
Now answer the questions based on data in table given above.
i) What is the radius of curvature of the lens ?
ii) Object distance 20 cm, and image distance is 60 cm. Find focal length.
iii) What is the object distance when a virtual image is formed at a distance of 30 cm ?
iv) For what distance of object, image is formed at infinite distance ?

Answer 1

ANSWER:

An object is placed 30 cm in front of convex lens of focal length 10 cm :

Given :

${u}^{i} = - 30cm \\ \\ {f}^{i} = - 10cm \\ \\ from \: lens \: formulae \\ \\ \frac{1}{ {v}^{i} } - \frac{1}{ {u}^{i} } = \frac{1}{ {f}^{i} } \\ \\ \frac{1}{ {v}^{i} } - \frac{1}{ - 30} = \frac{1}{10} \\ \\ {v}^{i} = 15cm$

Thus image is formed at a distance of 15 cm behind the lens.

Now a concave lens of focal length

${f}^{f}$

is placed in contact with convex lens. so the screen has to be shifted by 45 cm further away.

Now the new image distance

${v}^{f}$

=15+45=60 cm

Let the focal length of combination of lens be F.

Using lens formula,

$\frac{1}{ {v}^{f} } - \frac{1}{ {u}^{f} } = \frac{1}{ {f}^{} } \\ \\ \frac{1}{ {f}^{} } = \frac{1}{60} - \frac{1}{ - 30} \\ \\ {f}^{} = 20cm$

Focal length of combination of lenses

$\frac{1}{f} = \frac{1}{ {f}^{i} } + \frac{1}{ {f}^{f} } \\ \\ \frac{1}{20} = \frac{1}{10} + \frac{1}{ {f}^{f} } \\ \\ {f}^{f} = - 20cm$

=−20 cm (minus sign comes as the focal length of concave lens is negative)