Question

A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is
(a) a convex lens of focal length 20 cm, and
(b) a concave lens of focal length 16 cm?

Answer 1

Hint- In the given problem we have been given that the object is virtual because the lens is placed in the path of the convergent beam, and the point P will lie on the right of the lens and will act as a virtual object.
Then in this type of problems when lens is placed in the path of the convergent beam 12 cm from P, we will take this value as object distance,
So, object distance = $$u=12cm$$.

Step By Step Answer:
(a)Here the focal length of the convex lens is given as $=f_1=+20cm$

Now using the lens formula to get the image distance because this will be the point where the beam converges if the lens is convex lens

${\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}$

Now putting the given values of the focal length and object distance in the lens formula,

${\displaystyle \frac{1}{20}}={\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{12}}$
$\Rightarrow {\displaystyle \frac{1}{v}}={\displaystyle \frac{1}{20}}+{\displaystyle \frac{1}{12}}={\displaystyle \frac{3+5}{60}}={\displaystyle \frac{8}{60}}$
$\Rightarrow v={\displaystyle \frac{60}{8}}$$=7.5cm$

So, the beam converges at a point $7.5cm$from the lens and it is a real image.

(b)Now in this section of the problem the given lens is concave lens and the focal length of this concave lens$=f=-16cm$

Now again applying the lens formula for the given situation

${\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}$
$\Rightarrow {\displaystyle \frac{1}{-16}}={\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{12}}$
$\Rightarrow {\displaystyle \frac{1}{v}}={\displaystyle \frac{-1}{16}}+{\displaystyle \frac{1}{12}}$
$\Rightarrow {\displaystyle \frac{1}{v}}={\displaystyle \frac{-3+4}{48}}={\displaystyle \frac{1}{48}}$
$\Rightarrow v=48cm$

So, in this case we have got the image distance=48cm

Hence the beam of light will converge at the distance of 48cm from the lens. This is a real image.

Note-In both types of the lenses the image obtained is a real image, not a virtual one, that means the beam really converges at a point, it does not only seem to be converging at a point. So, both the lenses will make a real image in the given situation.