Let u and v be the distances of the object and the image from a lens of focal length f. The correct graphical representation of u and v for a convex lens when |u| > f, is

Geometrical Optics PYQ — JEE Main 2025 · 29 Jan · Shift 1 · Q63 with Solution

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Correct answer: D

Lens formula

For a thin convex lens, using Cartesian sign convention,

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

For a real object placed to the left of the lens, $u<0$ . The condition $|u|>f$ means the object is outside the focal length.

Rewrite in terms of positive object distance magnitude

Let $x=|u|=-u$ where $x>f$ . Then the lens formula becomes

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

So,

\frac{1}{v}=\frac{1}{f}-\frac{1}{x}=\frac{x-f}{fx}

Hence,

v=\frac{fx}{x-f}

Since $x=|u|$ , we can write

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

Find the relation between $u$ and $v$

Using $x=|u|$ ,

v=\frac{fx}{x-f}

This is a rectangular hyperbola.

Multiplying,

v(x-f)=fx

vx-fv=fx

x(v-f)=fv

So,

x=\frac{fv}{v-f}

Thus both $x=|u|$ and $v$ vary hyperbolically.

Important graphical features

For

v=\frac{fx}{x-f}, \quad x>f

the graph has:

Vertical asymptote: $x=f$
Horizontal asymptote: $v=f$
It passes through $x=2f \Rightarrow v=2f$
For $x\to f^+$ , $v\to \infty$
For $x\to \infty$ , $v\to f^+$

So the correct graph must be a decreasing curve in the first quadrant with asymptotes $|u|=f$ and $v=f$ .

If the graph is drawn using signed $u$ on the horizontal axis, then since $u<0$ , the curve lies in the second quadrant and has asymptotes $u=-f$ and $v=f$ .

Conclusion

Therefore, the correct graphical representation is the one corresponding to a hyperbola with the above asymptotic behavior. This matches Option D.

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