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)
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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