Science — CBSE Class 10 board question

(a) The magnification produced by a lens is given by:

$$m = \frac{v}{u}$$

where $v$ is the image distance and $u$ is the object distance from the optical centre.

(b) Given: Object height $h$ = 4 cm, $f$ = +20 cm, $u$ = −10 cm

Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$$\frac{1}{v} = \frac{1}{20} + \frac{1}{(-10)} = \frac{1}{20} - \frac{1}{10} = \frac{-1}{20}$$

So $v$ = −20 cm

Magnification: $m = \frac{v}{u} = \frac{-20}{-10} = +2$

Size of image: $h' = m \times h = 2 \times 4 = \textbf{8 cm}$

The image is virtual, erect and enlarged.

Source: Chapter 9, Section 9.3

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• (a) is a standard 1-mark definition — write only the formula with symbols defined.
• (b) Since $u$ = 10 cm < $f$ = 20 cm, the object is between F and O; expect a virtual, erect, magnified image (consistent with Table 9.4). Always use sign convention: distances measured against incident light are negative, so $u = -10$ cm.
• Show all steps clearly — substitution, calculation of $v$, then $m$, then $h'$ — each step can carry a mark.