Question

In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position; the screen is adjusted to get a clear image of the object. A graph between the object distance $u$ and the image distance $v$ , from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of ${45^ \circ }$ with the x-axis meets the experimental curve at P. The coordinates of P will be
(A) $\left( {2f,2f} \right)$
(B) $\left( {\dfrac{f}{2},\dfrac{f}{2}} \right)$
(C) $\left( {f,f} \right)$
(D) $\left( {4f,4f} \right)$

Answer 1

To answer this question, we have to use the lens formula to find the equations of the two curves given in the question, which are intersecting each other at the point P. Then, we need to solve those two equations to get the final coordinates of the point.

Formula used:
The formula used in solving this question is given by
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$ , where $f$ is the focal length of a lens, $v$ is the distance of the image from the centre of the lens, and $u$ is the distance of the object from the lens.

Complete step by step solution:
According to the question, the experimental curve is drawn between the image and the object distances from the convex lens. This means that the equation of the experimental curve is the same as the lens formula, which is given by
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$ ……………..… (1)
Now, as the lens is convex, so using the Cartesian sign convention for a convex lens, we have
$f = + f$ , $v = + v$ , and $u = - u$ . Substituting these in (1) we get the equation of the experimental curve as
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{{ - u}}$
Or
$\Rightarrow \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ ……………...… (2)
Now, we have to find the equation of the straight line. According to the question, the straight line is making an angle of ${45^ \circ }$ with the x-axis. Therefore, the slope of the straight line is
$\Rightarrow m = \tan {45^ \circ }$
$\Rightarrow m = 1$ …………………..… (3)
Also, the straight line is passing through the origin $\left( {0,0} \right)$
Therefore, the equation of the line is given by the point slope form as
$\Rightarrow \dfrac{{v - 0}}{{u - 0}} = m$
$\Rightarrow \dfrac{v}{u} = m$
From (3)
$\Rightarrow \dfrac{v}{u} = 1$
Or $v = u$ ………………...… (4)
Now, as the point P is the intersection point of the experimental curve and the straight line, so its coordinates are given by solving (2) and (3).
Substituting (4) in (3), we have
$\Rightarrow \dfrac{1}{u} + \dfrac{1}{u} = \dfrac{1}{f}$
$\Rightarrow \dfrac{2}{u} = \dfrac{1}{f}$
Taking reciprocal, we get
$\Rightarrow \dfrac{u}{2} = f$
$\Rightarrow u = 2f$
From (4)
$\Rightarrow v = u = 2f$
Thus, the coordinates of the point P are $\left( {2f,2f} \right)$ .
Hence, the correct answer is option (A).

Note:
Don’t be confused regarding the signs of the object and the image distances. In an optics experiment, the distances are always taken to be positive while being plotted on the graph. So $u$ and $v$ both are positive in this question.