Question

The focal length of a thin biconvex lens is $20cm$. When an object is moved from a distance of $25cm$ in front of it to $50cm$, the magnification of its image changes from ${m_{25}}$​ to${m_{50}}$​. The ratio of${m_{25}}$​ to ${m_{50}}$ is:
A) $2$
B) $4$
C) $6$
D) $8$

Answer 1

First we calculate distance of an image for given two distances of object as $25cm$ and $50cm$. Then we calculate magnification by formula used in case of distance of an object and an image.

Formula used:
We are using lens formula $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$ to calculate $v$ in both distance of an object $u$. Magnification is calculated by $m = \dfrac{{ - v}}{u}$.

Complete step by step solution:
Given: focal length of biconvex lens, $f = 20$, distance of an object $u = {\text{ }}25{\text{ }}cm$ and 50$$$$cm.
First we calculate the distance of the image for $u = {\text{ }}25{\text{ }}cm$.
We know that the lens formula is given as
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
$\Rightarrow \dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}$

$$\dfrac{1}{v} = \dfrac{1}{{20}} + \dfrac{1}{{( - 25)}} = \dfrac{1}{{20}} - \dfrac{1}{{25}} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{{5 - 4}}{{100}} = \dfrac{1}{{100}} \\\ v = 100{\text{ }}cm \\\$$

Hence, the distance of the object for $u = {\text{ }}25{\text{ }}cm$ is $v = 100{\text{ }}cm$.
Magnification for $u = {\text{ }}25{\text{ }}cm$ is given by, ${m_{25}} = \dfrac{{ - v}}{u}$

$${m_{25}} = \dfrac{{ - v}}{u} \\\ {m_{25}} = \dfrac{{ - 100}}{{ - 25}} = 4 \\\$$

Again we calculate distance of an image for
$\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}$

$$\dfrac{1}{v} = \dfrac{1}{{20}} + \dfrac{1}{{( - 50)}} = \dfrac{1}{{20}} - \dfrac{1}{{50}} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{{5 - 2}}{{100}} = \dfrac{3}{{100}} \\\ \therefore v = \dfrac{{100}}{3}{\text{ }}cm \\\$$

Hence, distance of an image for $u = - 50{\text{ }}cm$ is $v = \dfrac{{100}}{3}{\text{ }}cm$.
And magnification for is given by, ${m_{50}} = \dfrac{{ - v}}{u}$

$${m_{50}} = \dfrac{{ - \dfrac{{100}}{3}}}{{ - 50}} = \dfrac{{100}}{{50 \times 3}} \\\ {m_{50}} = \dfrac{2}{3} \\\$$

Hence, the ratio of ${m_{25}}$to ${m_{50}}$, i.e. ${m_{25}}:{m_{50}}$ is given by

$$\dfrac{{{m_{25}}}}{{{m_{50}}}} = \dfrac{4}{{\dfrac{2}{3}}} \\\ \Rightarrow \dfrac{{{m_{25}}}}{{{m_{50}}}} = 4 \times \dfrac{3}{2} = 6 \\\ \therefore {m_{25}}:{m_{50}} = 6 \\\$$

Hence, the required ratio is given by, ${m_{25}}:{m_{50}} = 6$

Therefore, the correct option is C.

Additional information:
The ratio of the height of an image to the height of an object is known as Magnification of a lens. It is also described as the ratio of image distance to the object distance. The distance of the object is $u$, the distance of the image is $v$. Then we calculate magnification of a lens by the formula $m = \dfrac{{ - v}}{u}$. In case of height: height of image=$h'$ and height of an object = $h$. Then, magnification of a lens is given by $m = \dfrac{{h'}}{h}$.

Note: We know that distance taken in the lens may be positive or negative. Students must be careful to choose the sign for $u$ and $v$. Students must choose negative signs for distance taken for objects, i.e., $u$.
Here magnification is calculated by only with $m = \dfrac{{ - v}}{u}$ not by $m = \dfrac{{h'}}{h}$. We can’t use $m = \dfrac{{h'}}{h}$ because in question the height of the image and object is not given. Students must choose the correct formula for calculating magnification.