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  • A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20cm, and (b) a concave lens of focal length 16cm?

Q 9.8  A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20cm, and (b) a concave lens of focal length 16cm?

Answers (1)

best_answer

In any Lens 

:\frac{1}{v} - \frac{1}{u}=\frac{1}{f} 

\\v= the distance of the image from the optical centre

\\u= the distance of the object from the optical centre

\\f= the focal length of the lens

a)

Here, The beam converges from the convex lens to point P. This image P  will now act as an object for the new lens which is placed 12 cm from it and focal length being 20 cm.

So

, \frac{1}{v} - \frac{1}{12}=\frac{1}{20}

    \frac{1}{v}=\frac{1}{20} + \frac{1}{12}

\frac{1}{v}=\frac{8}{60}

v = 7.5 cm

Hence distance of image is 7.5 cm and it will form towards the right as the positive sign suggests.

b)

HERE, Focal length f = -16cm

so,

\frac{1}{v} - \frac{1}{12}=\frac{1}{-16}

 

\frac{1}{v} =\frac{1}{-16} + \frac{1}{12} = \frac{1}{48}

v = 48 cm

Hence image distance will be 48 cm in this case, and it will be in the right direction(as the positive sign suggests)

Posted by

Pankaj Sanodiya

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