WAEC - Physics (2015 - No. 27)
An object is placed 15cm from a diverging lens of focal length 12cm. The image of the object formed by the lens in
real and 6.67cm from the lens
virtual and 6.67cm from the lens
real and 60.00cm from the lens
virtual and 60.00cm from the lens
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The focal length of a diverging lens is negative.
f = -12 cm; u = 15 cm.
\(f = \frac{uv}{u + v}\)
\(-12 = \frac{15v}{15 + v}\)
\(-12 (15 + v) = 15v \implies -180 - 12v = 15v\)
\(-180 = 15v + 12v \implies -180 = 27v\)
\(v = \frac{-180}{27} = -6.67 cm\)
The image is virtual and 6.67 cm from the lens.