Question

An object and a screen are mounted on an optical bench and a converging lens is placed between them so that a sharp image is received on the screen. The linear magnification of the image is $2.5$. The lens is now moved $30cm$ nearer the screen and a sharp image is again formed on the screen, the focal length of the lens is:
(A) $14.0cm$
(B) $14.3cm$
(C) $14.6cm$
(D) $14.9cm$

Answer 1

Hint : You can use the lens formula which is $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ in order to calculate and substitute in the further formulas. Linear magnification can be defined as the ratio of image distance to the distance of the object.

Complete step by Step Solution:
Convex lenses are often referred to as converging lenses because the beams meet after the convex lens is lowered, whereas the concave lenses are known as divergent lenses, as after the concentric lens, the Beams vary. Images created by these lenses may depend on the location and scale of the lenses and may be actual or simulated.
Firstly, using the lens formula,
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$
Where, $v$ is the image distance, $u$ is the object distance and $f$ is the focal length,
For position one, where the object and a screen are mounted on an optical bench and a converging lens is placed between them so that a sharp image is received on the screen
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ (Eq.1)
And for position two, where the lens is now moved $30cm$ nearer the screen and a sharp image is again formed on the screen,
$\dfrac{1}{{v - 30}} + \dfrac{1}{{u + 30}} = \dfrac{1}{f}$ (Eq. 2)
Now, it is given in the question that linear magnification is $2.5$
Now,
$m = \dfrac{v}{u}$
Where, $v$ is the image distance, $u$ is the object distance and $m$ is the linear magnification.
$\dfrac{v}{u} = 2.5$
$\Rightarrow v = 2.5u$
Now, substituting $v = 2.5u$ in the equation $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
$\dfrac{1}{{2.5u}} + \dfrac{1}{u} = \dfrac{1}{f}$
$\Rightarrow u = \dfrac{{7f}}{5}$ (Eq.3)
Equating equations one and two, we get
$\dfrac{1}{{v - 30}} + \dfrac{1}{{u + 30}} = \dfrac{1}{v} + \dfrac{1}{u}$
$\Rightarrow \dfrac{1}{{2.5u - 30}} + \dfrac{1}{{u + 30}} = \dfrac{1}{{2.5u}} + \dfrac{1}{u}$
Using equation 3, we get,
$\dfrac{1}{{2.5 \times (\dfrac{{7f}}{5})}} + \dfrac{1}{{\dfrac{{7f}}{5}}} = \dfrac{1}{{2.5 \times (\dfrac{{7f}}{5}) - 30}} + \dfrac{1}{{\dfrac{{7f}}{5} + 30}}$
Solving the above given equation, we get
$f = \dfrac{{100}}{7}$
$\Rightarrow f = 14.3cm$

Note: The intensity of a lens is the measuring of the degree at which the light rays land on it. The degree of convergence depends on the lens' focal length. We then define the lens 'power to be the inverse of the lens' focal length.