Question

An object is placed $$30cm$$ away from a convex lens of focal length $$10cm$$ and a sharp image is formed on a screen. Now a concave lens is placed in contact with the convex lens. The screen now has to be moved by $$45cm$$ to get a sharp image again. The magnitude of the focal length of the concave lens is (in cm).
$\left(A\right)72$
$\left(B\right)60$
$\left(C\right)36$
$\left(D\right)20$

Answer 1

Hint: The lens that converges rays of light that is parallel to its principal axis. This lens is called a convex lens. The lens that diverges rays of light is called a concave lens. For an object, distance finds the image distance using the lens formula. Now for the new object and image distance the focal length of the concave lens. Then find the focal length of the combination.

Formula used:
lens formula:
${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$

Complete step by step solution:
The lens converges rays of light that are parallel to its principal axis. This lens is called a convex lens. The lens that diverges rays of light is called a concave lens. The distance between the center of a lens where parallel rays converge or diverge is called the focal length.
When the object is at infinity, the image formed at the focus of the convex lens is real and inverted. When an object is at an imaginary point, then the image will be real, inverted, and of the same size. When an object is at the focal point then the image will be at infinity.
For lenses the sign convention:
The focal length of a convex lens is positive and for a concave lens is negative.
Optical center of a lens lies on the origin of the$$x-y$$axis.

Here it is given that object distance is $u_1=-30cm$, $f_1=10cm$
Then from the lens formula
${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{-30}}={\displaystyle \frac{1}{10}}\Rightarrow v=15cm$
Thus, the image is formed at $15cm$ behind the lens.
Here the concave lens of the focal length $f_2$ is placed in contact with the convex lens. Hence the screen is shifted by $45cm$ further away.
New image distance $v_2=15+45=60cm$
Then by using lens formula for combination of lenses
$\Rightarrow$ ${\displaystyle \frac{1}{60}}-{\displaystyle \frac{1}{-30}}={\displaystyle \frac{1}{F}}\Rightarrow F=20cm$
Then the focal length of the concave lens
$\Rightarrow$ ${\displaystyle \frac{1}{F}}={\displaystyle \frac{1}{f_1}}+{\displaystyle \frac{1}{f_2}}$
$\Rightarrow$ ${\displaystyle \frac{1}{20}}={\displaystyle \frac{1}{10}}+{\displaystyle \frac{1}{f_2}}$
$\therefore$ $f_2=-20cm$

Note: While you're solving problems related to lens, we should take proper care of the signs of different quantities. The focal length of a convex lens is positive and for the concave lens is negative. Optical center of a lens lies on the origin of the $$x-y$$ axis. The lens converges rays of light that are parallel to its principal axis.