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


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