Ray Optics And Optical Instruments Ques 32
32. 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
[2011 M]
(a) $-10$ $ cm$
(b) $20$ $ cm$
(c) $-30$ $ cm$
(d) $5$ $ cm$
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Answer:
Correct Answer: 32.(c)
Solution:
- (d)
By the lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$u=10$ $ cm$
$v=15$ $ cm$
$f=$ \frac{1}{2\pi\sqrt{LC}}
Substituting the values, we obtain
$\frac{1}{15}-\frac{1}{10}=\frac{1}{f}$
$\frac{10-15}{150}=\frac{1}{f} \quad $
$\therefore f=-\frac{150}{3}=-30$ $ cm$
BETA
