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
The correct answer is: B
Explanation
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.