A student measures the focal length of a convex lens by putting an object pin at a distance 'u' from the lens and measuring the distance 'v' of the image pin. The graph between 'u' and 'v' plotted by the student should look like

Geometrical Optics PYQ — JEE Main 2008 · Shift 0 · Q60 with Solution

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Correct answer: C

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

For the usual experimental graph, we take the magnitudes of object distance and image distance, so the relation becomes

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

where $u>f$ and $v>f$ for a real image.

\frac{1}{v} = \frac{1}{f} - \frac{1}{u}

v = \frac{fu}{u-f}

It is not a straight line.
As $u \to f^+$ , $v \to \infty$ so there is a vertical asymptote at $u=f$
As $u \to \infty$ , $v \to f$ so there is a horizontal asymptote at $v=f$
When $u=2f$ , $v=\frac{f(2f)}{2f-f}=2f$ so the curve passes through $(2f,2f)$ .
Also, $\frac{dv}{du} = -\frac{f^2}{(u-f)^2}<0$ so $v$ decreases as $u$ increases.

Therefore, the $u$ - $v$ graph is a rectangular hyperbola in the first quadrant, decreasing, with asymptotes $u=f$ and $v=f$ .
Among the given options, this corresponds to Option C.

Derived answer: $\boxed{\text{C}}$