Question

When an object is placed between the focus and the pole of a concave mirror, the image formed is:
A. Real, inverted and small
B. Virtual, inverted and small
C. Real, inverted and enlarged
D. Virtual, erect and enlarged

Answer 1

Try to geometrically determine how the ray diagram for such a situation would look. Start off with the fact that a concave mirror has its reflecting curved surface bent inwards, away from the object. Then use the fact that the incidence and reflected angles are equal about the normal to the mirror surface to trace out two rays, one from the top of the object striking the pole of the mirror, and the other reflecting off through the focus. This will give you a clear picture of what the image will look like, from which you can deduce the correct option.

Complete step-by-step answer:
Let us begin by understanding what a concave mirror is and how it works.
A concave mirror or a converging mirror is a curved mirror whose reflecting surface is recessing inwards, away from the incident light. Light rays converge at a focus point when they strike the mirror surface and get reflected back from the reflecting concave surface. This happens because of the differential reflection of light rays at different angles since the normal to the mirror is different at each point on the mirror, owing to its curvature.
Concave mirrors can produce both real and virtual images. They can be upright (if virtual) or inverted (if real). They can be behind the mirror (if virtual) or in front of the mirror (if real). They can also be magnified, minified or the same size as the object.
In any case, we can determine the position of the image with respect to the object by employing the method of ray tracing, where we utilize the law of reflection (which suggests that the angle of reflection is equal to the angle of incidence about the normal to the surface) to trace out the emergent ray path, and consequently the image location and size.
We first begin by drawing a ray from the top of the object to the surface vertex (the point where the optic axis meets the mirror surface) in accordance with the law of reflection. We then draw a second ray from the top of the object parallel to the optic axis which gets reflected in such a way that the reflected ray passes through the focal point. The point at which these two rays meet is the point corresponding to where the image is formed. The distance of this point from the optic axis gives the height of the image.

In this case we see that the image formed using this procedure is a virtual image that is oriented straight upwards and is much larger in size than the object.
Therefore, the correct option would be: D. Virtual, erect and enlarged.

So, the correct answer is “Option D”.

Note: Remember that, for a spherical mirror, we say that the image is real when the image is formed on the same side as that of the object and is called virtual when the image is on the other side of the mirror. When the orientation of the image remains unchanged from the object, we say the image is erect, else we say it is inverted, and if the size of the image is larger or smaller in comparison to the object, we say that the image is magnified or minified respectively.