In an optics experiment, with the position of the object fixed, a student varies the position of the concave mirror of focal length ‘f’ 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 mirror is plotted taking two pairs $(u_{1}, v_{1}) & (u_{2}, v_{2})$ as shown in figure then the co-ordinates of point ‘p’ will be

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$(3 f, 3 f)$

$(f, f)$

$(2 f, 2 f)$

$(\frac{f}{2}, \frac{f}{2})$

answer is B.

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Detailed Solution

Let us write the equation of two straight lines. Intersection will coordinate of point P

$v = \frac{v_{1}}{u_{1}} u + v_{1}$

$v = \frac{v_{2}}{u_{2}} u + v_{2}$

Solving together we will get,

$u (\frac{v_{1}}{u_{1}} - \frac{v_{2}}{u_{2}}) = v_{2} - v_{1}$ ...........(1)

Now, we will use mirror formula, $\frac{1}{f} = \frac{1}{v_{1}} + \frac{1}{u_{1}}$

$\Rightarrow \frac{v_{1}}{f} = 1 + \frac{v_{1}}{u_{1}}$

$\Rightarrow \frac{v_{1}}{u_{1}} = 1 - \frac{v_{1}}{f}$ ...................(2)

Similarly, $\frac{1}{f} = \frac{1}{v_{2}} + \frac{1}{u_{2}}$

$\Rightarrow \frac{v_{2}}{f} = 1 + \frac{v_{2}}{u_{2}}$

$\Rightarrow \frac{v_{2}}{u_{2}} = 1 - \frac{v_{2}}{f}$ ...........................(3)

Put (2) and (3) in (1), we get

$u = f$

Similarly, $v = f$

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