Use the mirror equation to deduce that:

(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.

(b) a convex mirror always produces a virtual image independent of the location of the object.

(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.

(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

[Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]

Given:

(a) For a concave mirror, the focal length (f) is negative.
Therefore f < 0
When the object is placed on the left side of the mirror, the object distance (u) is
negative.
Therefore u < 0
For image distance v, we can write the lens formula as:

…(1)

The object lies between f and 2f.

2f < u < f

⇒ 1/2f < 1/u < 1/f

⇒ −1/2f < −1/u < −1/f

0 …(ii)

Using equation (1), we get:

1/2f < 1/v < 0

1/v is negative, i.e., v is negative.

1/2f < 1v

2f > v

-v > -2f
Therefore, the image lies beyond 2f.

(b) For a convex mirror, the focal length (f) is positive.
Therefore f > 0
When the object is placed on the left side of the mirror, the object distance (u) is
negative.
Therefore u < 0
For image distance v, we have the mirror formula:

…(3)

Using equation (2), we can conclude that:

1/v < 0

v > 0

Thus, the image is formed on the back side of the mirror.
Hence, a convex mirror always produces a virtual image, irrespective of the object distance.

(c) For a convex mirror, the focal length (f) is positive.
Therefore f > 0
When the object is placed on the left side of the mirror, the object distance (u) is
negative,
Therefore u < 0
For image distance v, we have the mirror formula:

⇒

But we have u < 0

Therefore 1/v > 1/f

v < f
Hence, the image formed is diminished and is located between the focus (f) and the pole.

(d) For a concave mirror, the focal length (f) is negative.

Therefore f < 0

When the object is placed on the left side of the mirror, the object distance (u) is
negative.
Therefore u < 0
It is placed between the focus (f) and the pole.

Therefore f > u > 0

1/f < 1/u < 0

1/f − 1/u < 0
For image distance v, we have the mirror formula:

Therefore 1/v < 0

v > 0
The image is formed on the right side of the mirror. Hence, it is a virtual image.
For u < 0 and v > 0, we can write:

1/u > 1/v

v > u
Magnification, m = v/u > 1
Hence, the formed image is enlarged