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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Geometrical Optics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Geometrical Optics

As depicted in the figure, a concave mirror of radius R is placed...

As depicted in the figure, a concave mirror of radius RR is placed on a horizontal table. Water (refractive index =μ= \mu) is poured into the mirror up to a height hh. At what position should an object be placed such that its image forms on itself?
As depicted in the figure, a concave mirror of radius R is placed on a horizontal table. Water (refractive index = μ )

Options

  1. AR−hμ\dfrac{R - h}{\mu} above the water surface
  2. BR+hμ\dfrac{R + h}{\mu} above the water surface
  3. Cμ(R−h)\mu(R - h) above the water surface
  4. DRμ−h\dfrac{R}{\mu} - h above the water surface

Correct answer

A. R−hμ\dfrac{R - h}{\mu} above the water surface

Step-by-step solution

For the final image to form on the object itself, the light rays must retrace their path after reflection from the concave mirror.

This happens when the rays strike the mirror normally, which means the refracted rays must appear to come from the center of curvature CC of the mirror.

The center of curvature CC is at a distance RR from the pole of the mirror. Since the water is filled up to a height hh, the distance of CC above the water surface is R−hR - h.

Let the object be placed at a height xx above the water surface. When viewed from the water (denser medium), the apparent height of the object due to refraction at the plane surface is given by:
Apparent height =μx= \mu x

For the rays to appear to come from CC, this apparent height must be equal to the distance of CC from the water surface:
μx=R−h\mu x = R - h

x=R−hμx = \dfrac{R - h}{\mu}

Thus, the object should be placed at a distance of R−hμ\dfrac{R - h}{\mu} above the water surface.
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