Looking for classes? Ksquare Career Institute, Bengaluru →
A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point 15 cm from the lens on the opposite side. If the lens is removed the point where the rays meet will move 5 cm closer to the lens. The focal length of the lens is
A
−30 cm
B
5 cm
C
−10 cm
D
20 cm
Detailed Solution
Without the lens the converging rays would meet at a point 15 − 5 = 10 cm behind the lens position. This point acts as a virtual object for the lens.
For a virtual object on the far side of the lens, u = +10 cm.
With the lens the rays actually meet 15 cm from the lens on the opposite side, so v = +15 cm.
Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{f} = \frac{1}{15} - \frac{1}{10} = \frac{2 - 3}{30} = -\frac{1}{30}$
f = −30 cm (negative, as expected for a diverging lens)
For a virtual object on the far side of the lens, u = +10 cm.
With the lens the rays actually meet 15 cm from the lens on the opposite side, so v = +15 cm.
Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{f} = \frac{1}{15} - \frac{1}{10} = \frac{2 - 3}{30} = -\frac{1}{30}$
f = −30 cm (negative, as expected for a diverging lens)