The problem involves a mirror forming a virtual image, which means the mirror is a concave mirror. The magnification (m) of the mirror is given by the ratio of the image height to the object height. In this case, the image is reduced to one-third of the object size, so m = -1/3 (the negative sign indicates that the image is virtual).

The mirror formula, which relates the object distance (u), the image distance (v), and the focal length (f) of the mirror, is given by:

1/f = 1/v + 1/u

We know that m = -v/u. We can rearrange this to find v = -mu.

Substituting v into the mirror formula gives:

1/f = -1/mu + 1/u

Rearranging terms gives:

1/f + 1/mu = 1/u

Substituting the given values f = 25 cm and m = -1/3 into this equation gives:

1/25 + 1/(-1/3 * u) = 1/u

Solving this equation for u gives u = 75 cm.

So, the correct answer is C. 75 cm.

Question

The problem involves a mirror forming a virtual image, which means the mirror is a concave mirror. The magnification (m) of the mirror is given by the ratio of the image height to the object height. In this case, the image is reduced to one-third of the object size, so m = -1/3 (the negative sign indicates that the image is virtual).

The mirror formula, which relates the object distance (u), the image distance (v), and the focal length (f) of the mirror, is given by:

1/f = 1/v + 1/u

We know that m = -v/u. We can rearrange this to find v = -mu.

Substituting v into the mirror formula gives:

1/f = -1/mu + 1/u

Rearranging terms gives:

1/f + 1/mu = 1/u

Substituting the given values f = 25 cm and m = -1/3 into this equation gives:

1/25 + 1/(-1/3 * u) = 1/u

Solving this equation for u gives u = 75 cm.

So, the correct answer is C. 75 cm.

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