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Question

An object 3 cm high is placed perpendicular to the principal axis of a concave lens of focal length 15 cm. The image is formed at a distance of 10 cm from the lens. Calculate a. The distance of object placed b. size and nature of the image formed.

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Solution

(a)

Lens formula:


1f=1v-1u1-15=1-10-1u1u=1-10+115 =-5150u=-30 cm

Hence, the distance of the object is 30 cm.

(b)

Magnification:

m=vu =-10-30 =13

Size of the image:

hi=mho =133 =1 cm

Here, the positive magnification implies erect and virtual image.

As the magnification is less than 1, the image is smaller than the object.




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