For a concave lens of focal length f, the relation between object and image distances u and v, respectively, from its pole can best be represented by (u = v is the reference line) :

Solution

Physics sir · 3 steps

A concave lens always forms the image closer than the object, so the curve must stay below u = v.

1. Lens formula. With signs, u = -x and f = -F, where x and F are the distances. Then 1/v = 1/f + 1/u = -1/F - 1/x, so the image is virtual and on the object's side.

   |v| = xF/(x + F)

   Object distance	Image distance	What the graph must show
   very small (x ≪ F)	|v| ≈ x	curve starts at the origin, running along the u = v line
   x = F	|v| = F/2	curve is below the u = v line, halfway up to f
   very large (x → ∞)	|v| → F	curve levels off at the dotted line v = f and never crosses it

2. Check the graphs. Since F/(x + F) < 1, we always get |v| < x, so the curve never rises above u = v. Since x/(x + F) < 1, we also get |v| < F, so the curve never rises above f.
3. So. Only graph (d) starts along u = v, stays below it, and flattens out at v = f. The answer is (d).

The idea to keep. For a diverging lens, the virtual image always lies between the lens and its focus: |v| < |u| and |v| < |f|. Test any graph at u → 0, u = f and u → ∞.

Answer from the NTA final answer key. Solution by PrepWiser.