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

Question

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) 5 cm (B) -10 cm (C) 20 cm (D) -30 cm

Tardigrade Admin · Accepted Answer

Here, v=+15 cm , u=+(15-5)=+10 cm According to lens formula (1/v)-(1/u)=(1/f) ⇒ (1/15)-(1/10)=(1/f) ⇒ f=-30 cm

Q. A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point $15 c m$ from the lens on the opposite side. If the lens is removed the point where the rays meet will move $5 c m$ closer to the lens. The focal length of the lens is

5 cm

-10 cm

20 cm

-30 cm

Here, $v = + 15 c m,$
$u = + (15 - 5) = + 10 c m$
According to lens formula
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\Rightarrow \frac{1}{15} - \frac{1}{10} = \frac{1}{f}$
$\Rightarrow f = - 30 c m$

Q. A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point 15cm from the lens on the opposite side. If the lens is removed the point where the rays meet will move 5cm closer to the lens. The focal length of the lens is