Science — CBSE Class 10 board question

(a) Focal length = 20 cm

In Observation 4, u = –40 cm and v = +40 cm, i.e., the image is formed at 2F on the other side. This means 2F = 40 cm, so f = 20 cm.
(Alternatively, applying lens formula to any observation gives f = +20 cm.)

(b) For Observation 3: u = –30 cm, v = +60 cm

$$m = \frac{v}{u} = \frac{+60}{-30} = \mathbf{-2}$$

The magnification is –2 (image is real, inverted and enlarged, twice the size of the object).

(c) In Observation 1, u = –15 cm (object between F and O), v = –60 cm (negative).
In Observation 2, u = –25 cm (object between F and 2F), v = +100 cm (positive).
Numerical magnification = 4 in both cases.

| | Observation 1 | Observation 2 |
|---|---|---|
| Nature | Virtual and erect | Real and inverted |
| Position | Same side as object | Opposite side of lens |

Source: Light – Reflection and Refraction, Sections 9.3.4, 9.3.7

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• (a) When |u| = |v|, the object is at 2F, so the image forms at 2F on the other side — this is the quickest way to read f without calculation. Observation 4 directly gives this.
• (b) Use m = v/u with correct signs. A negative m means real and inverted.
• (c) The key contrast is virtual/erect vs real/inverted, and image on same side vs opposite side. "Numerical magnification same" means |m| = 4 in both, but the sign (and hence nature) differs. Examiners expect two clear, distinct differences stated as a comparison.