An object is placed 15cm from a diverging lens of focal length 12cm. The image of the object formed by the lens in
-
A.
real and 6.67cm from the lens -
B.
virtual and 6.67cm from the lens -
C.
real and 60.00cm from the lens -
D.
virtual and 60.00cm from the lens
Correct Answer: Option 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.