Question

What is the focal length of a convex lens of focal length 30cm in contact with a concave lens of focal length 20cm? Is the system a converging or a diverging lens? Ignore the thickness of the lenses.

Answer 1

Hint – In this question use the concept that the focal length of the convex lens is always negative and the focal length of a concave lens is positive. Use the direct formula for the focal length of combined system that is ${\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{f_1}}+{\displaystyle \frac{1}{f_2}}$ where $f_1$ the focal length of the convex lens is and $f_2$ is the focal length of the concave lens.

Formula used: ${\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{f_1}}+{\displaystyle \frac{1}{f_2}}$.

Complete step-by-step solution-
Given data:
Focal length of the convex lens = 30cm
As we all know that the focal length of the convex lens is always positive as it behaves as a converging lens because it causes the light ray to converge at one point from the axis called as principal focus of the lens.
Let it is denoted by $f_1$
Therefore, $f_1$ = 30 cm.
Focal length of the concave lens = 20 cm.
As we all know that the focal length of the concave lens is always negative as it behaves as a diverging lens because it causes the light ray to bend away from the axis.
Let it is denoted by $f_2$
Therefore, $f_2$ = -20 cm.
Now we have to find out the focal length of the combined system i.e. a system of convex lens having focal length 30 cm and concave lens having focal length 20 cm, assuming negligence thickness of the lens.
Let the focal length of the combined system be f cm.
Therefore the focal length of the combined system is given as
${\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{f_1}}+{\displaystyle \frac{1}{f_2}}$
Now substitute the values we have,
 $\Rightarrow {\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{30}}+{\displaystyle \frac{1}{-20}}$
$\Rightarrow {\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{30}}-{\displaystyle \frac{1}{20}}$
Now simplify this by taking L.C.M we have,
$\Rightarrow {\displaystyle \frac{1}{f}}={\displaystyle \frac{20-30}{30\left(20\right)}}={\displaystyle \frac{-10}{600}}={\displaystyle \frac{-1}{60}}$
$\Rightarrow f=-60$cm
So as we see that the focal length comes negative so it behaves as a diverging lens.
So this is the required answer.

Note – In real life convex lenses are used to solve the problem of hypermetropia, since convex lenses are converging lenses therefore the rays from the object coming from a very far distance are converged towards the retina of the person wearing glasses made up of convex lens.