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  • A converging lens which has a focal length of 20 cm is placed 60 cm to the left of a concave mirror of focal length 30 cm. An object is placed 40 cm to the left of lens. The position of the final i

A converging lens which has a focal length of 20 cm is placed 60 cm to the left of a concave mirror of focal length 30 cm. An object is placed 40 cm to the left of lens. The position of the final image is:

Option: 1

30 cm


Option: 2

50 cm


Option: 3

60 cm


Option: 4

90 cm


Answers (1)

best_answer

For refraction through the lens

\mathrm{u=-40 \mathrm{~cm}, f=+20 \mathrm{~cm}}
\mathrm{ \text { Lens equation is : } \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \text { or } \frac{1}{v}-\frac{1}{-40}=\frac{1}{20} \text { or } \frac{1}{v}=\frac{1}{20}-\frac{1}{40}}
\mathrm{ or \, \mathrm{v}=+40 \mathrm{~cm}}

This gives the position of image \mathrm{ I_{1}}. Image \mathrm{ I_{1}} will be a real inverted image formed at 40 \mathrm{~cm} to the right of the lens. As concave mirror is placed to the right of lens at 60 \mathrm{~cm},hence the image I_{1}  is at a distance of 20 \mathrm{~cm} from the mirror. Therefore, image I_{1} becomes a real object for the concave mirror.

For reflection by concave mirror :
The position of the object I_{1} is within the focal length of the concave mirror Here, \mathrm{u=-20 \mathrm{~cm}, f=-30 \mathrm{~cm}}

Mirror formula is :
\mathrm{\frac{1}{v}+\frac{1}{u}=\frac{1}{f} \quad\, or \, \quad \frac{1}{v}-\frac{1}{20}=-\frac{1}{30}}
\mathrm{or \frac{1}{v}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60} \, or\, v=+60 \mathrm{~cm}}

This gives the position of final image \mathrm{I}.

Posted by

manish

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