An object of height $1 cm$ is kept perpendicular to the principal axis of a convex mirror of radius of curvature $20 cm$ . Distance of the object from the mirror is $20 cm$ . If the distance between heads of the image and the object will be $\sqrt{\frac{1601 \times x}{y}}$ cm, then what is the value of $y - x$ .

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Solution

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$
$\Rightarrow \frac{1}{v} = \frac{1}{10} - \frac{1}{- 20}$
$\Rightarrow v = \frac{20}{3} cm$
$I = - \frac{v}{u} \times O = \frac{\frac{20}{3}}{20} \times 1 = \frac{1}{3}$

$S = \sqrt{{(20 + \frac{20}{3})}^{2} + {(1 - \frac{1}{3})}^{2}} = \sqrt{\frac{6404}{9}} cm$

Why this question?

To learn to use the mirror formula and to find out the image size when the object size is known.

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An object of height 1 cm is kept perpendicular to the principal axis of a convex mirror of radius of curvature 20 cm. Distance of the object from the mirror is 20 cm. If the distance between heads of the image and the object will be √1601×xy cm, then what is the value of y−x.

1v=1f−1u ⇒1v=110−1−20 ⇒v=203 cm I=−vu×O=20320×1=13 S=√(20+203)2+(1−13)2=√64049 cm Why this question? To learn to use the mirror formula and to find out the image size when the object size is known.