A student plots a graph between inverse of magnification \[\dfrac{1}{m}\]produced by a convex thin lens and the object distance is as shown in figure. What was the focal length of the lens used?

A)\[\dfrac{b}{{ca}}\]
B)\[\dfrac{{bc}}{a}\]
C)\[\dfrac{c}{b}\]
D)\[\dfrac{b}{c}\]

Hint: Magnification of a lens is defined as the ratio of the height of image to the height of object. It is also written in terms of image distance and object distance. It is used to find the nature and the height of the image formed by the lens.
\[m = \dfrac{{{h_i}}}{{{h_o}}} = \dfrac{v}{u}\]
Where m = magnification of the lens
\[{h_i}\]=height of image
\[{h_o}\]=height of object

Step-by-Step Explanation:
Step I:
Given the slope of the curve is \[m = \dfrac{b}{c}\]
Also the lens formula is given by\[\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\]……(i)
Step II:
Multiply both sides of equation (i) by ‘u’, the equation becomes\[\dfrac{u}{v} - 1 = \dfrac{u}{f}\]……(ii)
Now it is known that magnification is given by\[m = \dfrac{v}{u}\]
Step III:
Substitute value of m in equation (ii),
\[\dfrac{1}{m} - 1 = \dfrac{u}{f}\]
\[\dfrac{1}{m} = \dfrac{u}{f} + 1\]
Or it can be written as
\[\dfrac{1}{m} = (\dfrac{1}{f})u + 1\]……(iii)
Step IV:
Compare it with the equation of a straight line \[y = mx + c\]…….(iv)
Comparing (iii) and (iv), evaluate the value of m,\[m = \dfrac{1}{f}\]
But ‘m’ is the given slope and the value of slope is already given.
\[\dfrac{1}{f} = \dfrac{b}{c}\]
\[f = \dfrac{c}{b}\]
Step V: The focal length of the lens is \[f = \dfrac{c}{b}\]

Option C is the right answer.

Note:Magnification is the ability of the lens to how much extent it can enlarge the image. Also it tells the nature of the image formed. Since convex lenses form both real and virtual images, magnification of the lens can be positive or negative depending on the nature of the image.
For instance if the value of magnification is positive, then if the position of the object is erect the image formed will also be erect. If the position of the object is inverted, then an inverted image will be formed.
But if the value of magnification is negative, then the convex lens will form a real image.