Where should the object be placed around the concave mirror to ensure that the size of the image is the same a

Question

Where should the object be placed around the concave mirror to ensure that the size of the image is the same as the size of the object?

Examoogle · Accepted Answer

The correct answer is Option C: At C, centre of curvature. For a concave mirror, an image of the same size as the object is formed only when the object is placed at the centre of curvature ($C$). At this position, the image is real, inverted, and located at the same point ($C$).

Step-by-Step Solution & Concepts:
For a concave mirror, an image of the same size as the object is formed only when the object is placed at the centre of curvature ($C$). At this position, the image is real, inverted, and located at the same point ($C$).

Working Principle: When an object is placed at $C$, a light ray passing parallel to the principal axis passes through the focus ($F$) after reflection, and a second ray passing through the focus ($F$) emerges parallel to the principal axis. The intersection of these two reflected rays occurs exactly at $C$, resulting in an image with magnification ($m$) of $-1$. The negative sign indicates the image is inverted, and the magnitude of 1 indicates the size is equal to that of the object.

Key Formulas: $m = \frac{h_i}{h_o} = -\frac{v}{u}$ — where $m$ is magnification, $h_i$ is image height, $h_o$ is object height, $v$ is image distance, and $u$ is object distance.; $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ — the mirror formula relating focal length ($f$), image distance ($v$), and object distance ($u$).

Key Points: Image at $C$ is real and inverted.; The distance of the object from the pole is $2f$ (equal to the radius of curvature $R$).; Magnification is exactly $-1$ at this position.

Advantages: Provides a clear method for determining the focal length of a concave mirror by finding the point where image size equals object size.

Applications: Used in laboratory experiments to verify focal length.; Used in optical arrangements where 1:1 image projection is required.

Additional Info: At other positions: beyond $C$ gives a diminished image, and between $C$ and $F$ gives a magnified image.. Option A is incorrect because between $F$ and $C$ results in an enlarged image. Option B is incorrect as an object at the focus results in the image being formed at infinity. Option D is incorrect as placing an object between $F$ and $P$ creates a virtual, magnified, and erect image.

Answer: C is correct — Placing the object at the centre of curvature ($C$) of a concave mirror produces a real, inverted image of the same size as the object.

Exam Tip: Remember that for all spherical mirrors and lenses, any special ray passing through the centre of curvature ($C$) reflects along the same path, confirming the location of the image.

Examoogle · Answer

Option A: Between focus F and centre of curvature C

Examoogle · Answer

Option B: At focus F

Examoogle · Answer

Option D: Between focus F and pole P

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