Observation table for a convex lens:
S.No. | Object distance (u) cm | Image distance (v) cm
1 | –150 | +30
2 | –75 | +37.5
3 | –50 | +50
4 | –37.5 | +75
5 | –30 | +150
6 | –15 | +37.5
The variation of image distance (v) with the object distance (u) for a convex lens is given in the following observation table. Analyse it and answer the questions given below :
Generated by claude-sonnet-4-6 · 2026-06-15 06:48 · grounding stimulus
Model Answer
(a) Focal length = 25 cm.
At S.No. 3, u = –50 cm and v = +50 cm, meaning the object is at 2F and the image is also at 2F. This happens only when u = 2f, so 2f = 50 cm → f = 25 cm.
(b) Observation S.No. 6 (u = –15 cm) is incorrect. Since u = –15 cm is less than f (25 cm), the object lies between F and the lens; a convex lens then forms a virtual image on the same side as the object (v should be negative), but the table shows v = +37.5 cm (positive/real), which is wrong.
(c)
$$m = \frac{v}{u} = \frac{+150}{-30} = -5$$
The image is real, inverted, and highly magnified (|m| > 1, m is negative). Negative magnification indicates a real and inverted image formed on the other side of the lens.
Source: Light – Reflection and Refraction, Convex Lens / Lens Formula section
---
Explanation
- Part (a): The quickest no-calculation clue is S.No. 3 where |u| = |v| = 50 cm. For a convex lens, u = v only when both equal 2f, giving f = 25 cm.
- Part (b): When the object is inside the focal length (u < f), a convex lens always produces a virtual, erect image — v must be negative in the sign convention used here. The table's positive v for S.No. 6 contradicts this.
- Part (c): Use m = v/u. Negative m → real + inverted; |m| = 5 → highly magnified. Always state all three characteristics of the image.