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  • A screen is placed a distance 40cm away from an illuminated object. A converging lens is placed between the source and the screen and it is attempted to form the image of the source on the screen.

A screen is placed a distance 40cm away from an illuminated object. A converging lens is placed between the source and the screen and it is attempted to form the image of the source on the screen. If no position could be found, the focal length of the lens

Option: 1

must be less than 10cm


Option: 2

must be greater than 10cm


Option: 3

must not be greater than 20cm


Option: 4

must not be less than 10cm.


Answers (1)

best_answer

Let the distance between source and lens be d. If image is formed on screen then v= 40 -d

Using lens equation  \mathrm{\frac{1}{v}-\frac{1}{u}=\frac{1}{f}}  we have
\mathrm{ \frac{1}{40-d}-\frac{1}{-d}=\frac{1}{f} \Rightarrow f=d-\frac{d^2}{40} }
For ' f ' to be maximum :  \mathrm{ \frac{d f}{d d}=0 }

\mathrm{ \Rightarrow \quad 1-\frac{d}{20}=0 \quad \Rightarrow d=20 \mathrm{~cm} \\ }

\mathrm{ \therefore \quad f_{\max }=20-\frac{(20)^2}{40}=10 \mathrm{~cm} }

This is maximum value of focal length for which image is formed on screen. But, in question image is not formed.

\mathrm{ \therefore f>10 \mathrm{~cm} }.
 

Posted by

manish painkra

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