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Standard XII

Physics

Magnification by Thin Lenses

A point objec...

Question

A point object $P$ is placed at center of curvature of a concave mirror of focal length $25 c m$ . The mirror is cut into two halves and shifted symmetrically $1 c m$ apart in perpendicular to the optical axis. The distance between images formed by both parts is:

2 c m

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1 c m

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3 c m

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4 c m

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Solution

The correct option is D $2 c m$
The image will be formed at the centre of curvature
The magnification of image will be $M = 1$
Therefore, $\Rightarrow h_{i} = h_{0}$

After separation from the axis, the distance between the images will be:
$\Rightarrow 1 \times 2$
$= 2 c m$

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A point object P is placed at center of curvature of a concave mirror of focal length 25cm. The mirror is cut into two halves and shifted symmetrically 1cm apart in perpendicular to the optical axis. The distance between images formed by both parts is:

The correct option is D 2cmThe image will be formed at the centre of curvature The magnification of image will be M=1Therefore, ⇒hi=h0After separation from the axis, the distance between the images will be:⇒1×2=2cm

A point object $P$ is placed at center of curvature of a concave mirror of focal length $25 c m$ . The mirror is cut into two halves and shifted symmetrically $1 c m$ apart in perpendicular to the optical axis. The distance between images formed by both parts is:

The correct option is D $2 c m$
The image will be formed at the centre of curvature
The magnification of image will be $M = 1$
Therefore, $\Rightarrow h_{i} = h_{0}$

After separation from the axis, the distance between the images will be:
$\Rightarrow 1 \times 2$
$= 2 c m$