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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.]

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

∴ f < 0 

When the object is placed on the left side of the mirror, the object distance (u) is negative. 

∴ u < 0 

For image distance v, we can write the lens formula as:

Therefore, the image lies beyond 2f.

 ( b)For a convex mirror, the focal length (f) is positive. 

∴ f > 0

 When the object is placed on the left side of the mirror, the object distance (u) is negative. 

∴ u < 0

 For image distance v, we have the mirror formula:

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

(c) For a convex mirror, the focal length (f) is positive. ∴ f > 0

 When the object is placed on the left side of the mirror, the object distance (u) is negative,

 ∴ u < 0

 For image distance v, we have the mirror formula:

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. 

∴ f < 0

When the object is placed on the left side of the mirror, the object distance (u) is negative.

 ∴ u < 0

 It is placed between the focus (f) and the pole.

For image distance v, we have the mirror formula:

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:

Hence, the formed image is enlarged.

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