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Subjects
Science CBSE
Class 10
Light reflection and refraction
Textbook questions

12. Exercises XII

Exercise condition:

5 m.

An object of size

$7.0\ cm$ is placed at

$27\ cm$ in front of a concave mirror of focal length

$18\ cm$ . At what distance from the mirror should a screen be placed so that a sharply focused image can be obtained? Find the size and nature of the image.

Mirror formula is,

$\frac{1}{v}\ -\ \frac{1}{u}\ =\ \frac{1}{f}$
$\frac{1}{v}\ +\ \frac{1}{u}\ =\ \frac{1}{f}$
$\frac{1}{u}\ =\ \frac{1}{v}\ +\ \frac{1}{f}$
$\frac{1}{v}\ =\ \frac{1}{f}\ +\ \frac{1}{u}$

Magnification is,

$M\ =\ \frac{-u}{v}$
$M\ =\ \frac{v}{u}$
$M\ =\ \frac{-v}{u}$
$M\ =\ \frac{u}{v}$

Image distance (

$v$ )

$=$

$cm$

Magnification (

$M$ )

$=$

Height of the image (

$h_i$ )

$=$

$cm$

Nature of the image is,

virtual, erect and enlarged size
virtual, erect and diminished size
real, inverted and diminished size
real, inverted, and enlarged size

Important!

Note: Enter your answers with no decimal digits.

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