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A 1 cm long awl pin is placed vertically in front of a concave mirror. A 0.5 cm long image of the awl pin is formed at 30 cm in front of the mirror. At what distance is the pin kept?

(Hint: magnification = I/O, Magnification = - v/u therefore, I/O = -v/u) Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

Answer is (b) – 20 cm

Explanation:

Here, size of object = O = + 10.0 mm = + 1.0 cm (as, 1 cm = 10 mm)

Size of Image size = I = 5.0 mm = 0.5 cm

Image distance, v = − 30 cm (as image is real)

Let, object distance = u

Focal length, f =?

Magnification m= I (Size of image)/

O(Size of object)

Magnification is given by m=−𝑣/𝑢

𝐼/𝑂= −𝑣/𝑢

0.5/1=−30/𝑢

U= -60cm

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