A student measures the focal length of 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

For a thin convex lens the lens formula is

$$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\qquad -(1)$$

In the common laboratory experiment the student measures only the magnitudes of the object distance $$u$$ and the image distance $$v$$, both taken as positive numbers. (The object is kept on the left of the lens, the real image forms on the right.)

Re-arrange $$-(1)$$ to express $$v$$ in terms of $$u$$:

$$\frac{1}{v}=\frac{1}{f}-\frac{1}{u} \;\Longrightarrow\; v=\frac{uf}{u-f}\qquad -(2)$$

Equation $$-(2)$$ describes the $$u\,$$-$$\,v$$ relation that will appear on the graph. Its salient features are:

• When the object approaches the focal length from the right, i.e. $$u\rightarrow f^{+}$$, the denominator $$u-f\rightarrow 0^{+}$$, hence $$v\rightarrow +\infty$$. Thus the curve has a vertical asymptote at $$u=f$$.

• When the object is very far from the lens ($$u\rightarrow\infty$$), $$\tfrac{f}{u}\rightarrow 0$$ and $$v\rightarrow f$$. Therefore the curve has a horizontal asymptote at $$v=f$$.

• For all admissible points ($$u\gt f$$) the derivative obtained from $$-(2)$$ is

$$\frac{dv}{du}=\frac{-f^{2}}{(u-f)^{2}}\lt0,$$

so the curve is monotonically decreasing throughout the first quadrant segment used in the experiment.

Hence the correct plot is a branch of a rectangular hyperbola that

• shoots up to infinity as $$u$$ approaches $$f$$ from the right, and
• gradually comes down, approaching the horizontal line $$v=f$$ as $$u$$ becomes very large.

Among the four sketches given in the problem, only Option C shows a decreasing hyperbola with a vertical asymptote at $$u=f$$ and a horizontal asymptote at $$v=f$$.

Therefore the required graph is the one in Option C.

Final Answer : Option C which is the hyperbolic curve described above.