Single Choice

An object is placed at a distance $$12\ cm$$ from a concave mirror of focal length $$6\ cm$$. The object moves $$2\ cm$$ towards the mirror. Find the distance moved by image.

A$$4\ cm$$
B$$3\ cm$$
Correct Answer
C$$5\ cm$$
D$$6\ cm$$

Solution

Initially the object was at center of curvature and we know image of object kept on the center of curvature forms on the center of curvature itself, so the image will be formed at $$-12\ cm$$
$$v_{\text{initial}} = -12\ cm$$

For final position as the object is moved towards the mirror $$u=-10\ cm$$
$$ \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$$

$$\dfrac{1}{v} + \dfrac{1}{(-10)} = \dfrac{1}{-6}$$

$$\Rightarrow\ \boxed{v = -15\ cm}$$

So distance moved by image will be
$$ = |15| - |12|$$

$$ = \boxed{3\ cm}$$

SIMILAR QUESTIONS

Optics

The speed at which the image of the luminous point object is moving, if the luminous point object is moving at speed $$v_0$$ towards a spherical mirror, along its axis, is: (Given: $$R=$$ radius of curvature, $$u=$$ object distance)

Optics

For which position of the object does a concave mirror produce an inverted, magnified and real image ?

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