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

Solution

Physics sir · 3 steps

Write v in terms of u and check the two ends: v blows up near f and settles down to f far away.

1. Lens formula. For a convex lens with |u| > f the image is real. In magnitudes the lens formula 1/v-1/u=1/f becomes 1/|v|+1/|u|=1/f.

   |v| = (f |u|)/(|u|-f) (|u|-f)(|v|-f) = f²

2. Check the ends. As |u| comes down towards f, the denominator goes to zero and |v| → ∞, so the dashed line |u| = f is a vertical asymptote. As |u| → ∞, |v| → f from above, so the curve flattens out but never touches the u-axis. In between, |u| = 2f gives |v| = 2f.
3. So. The graph is a rectangular hyperbola: v falls steadily as |u| grows, shooting up near the dashed line and levelling off far away. That is exactly graph (d).

The idea to keep. For real images from a convex lens, (|u|-f)(|v|-f)=f² is a rectangular hyperbola with asymptotes |u|=f and |v|=f. Always check u → f and u → ∞ before picking a graph.

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