1. Use the lens formula

For a thin lens,

For a concave lens, the focal length is negative. Let its magnitude be $f>0$, so actual focal length is $-f$. Hence,

2. Rearrange to get relation between $u$ and $v$

Multiply by $uvf$:

So,

This is the required relation between $u$ and $v$.

3. Nature of image for a concave lens

For a real object placed to the left of a concave lens:

• $u<0$
• image is virtual and on the same side, so $v<0$

Thus the physically relevant curve lies in the third quadrant of the $u$-$v$ plane.

4. Check asymptotic/limiting behavior

From

we get

Now examine limits:

• As $u\to 0^-$, $v\to 0^-.$
• As $u\to -\infty$, $v\to -f.$

So the graph starts near the origin in the third quadrant and approaches the horizontal line $v=-f$ as $u\to -\infty$.

5. Compare with the reference line $u=v$

For a concave lens, the image always lies between optical center and focus, so $|v|<|u|.$ Since both $u$ and $v$ are negative, this means $v>u.$ Therefore the curve lies above the line $u=v$ in the third quadrant.

6. Hence the correct graph

The correct option must show:

• branch in the third quadrant,
• passing near the origin,
• lying above the line $u=v$,
• approaching $v=-f$ for large negative $u$.

This corresponds to Option D.