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**Hint:**Use the mirror formula that gives a relation u, v and f and find an expression for v in the terms of u and f. Then find the relation between v and u at point P. Take a value of v greater than u and find the relation between v and f.

**Formula used:**

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

**Complete answer:**

Here, u is the position of the object and v is position of the image with respect to the mirror.

f is the focal length of the mirror.

The relation between u, v and f is given by $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$.

$\Rightarrow \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

$\Rightarrow \dfrac{1}{v}=\dfrac{u-f}{uf}$

$\Rightarrow v=\dfrac{uf}{u-f}$ ….. (i).

From the graph of v v/s u we can understand that as the object comes towards the mirror (as we reduce the object distance), the image distance increases. This means that v is inversely proportional to u.

Let us analyse the relation between u and v at point P. In the graph we can see that the point P also lies on a line passing through the origin and making an angle of ${{45}^{\circ }}$ with the positive x-axis.

From geometry we know that the x and y coordinates of the points on a line passing through the origin and making an angle of ${{45}^{\circ }}$ with the positive x-axis are equal.

This means that at point P, v = u.

Now, since the v is inversely proportional to u and at point P v = u, the value of v will be greater than the value of u for the points above point P.

Therefore, let v = nu, where n is a real number greater than one.

$\Rightarrow u=\dfrac{v}{n}$.

Substitute this value of u in (i).

$\Rightarrow v=\dfrac{\dfrac{v}{n}f}{\dfrac{v}{n}-f}$

$\Rightarrow \dfrac{v}{n}-f=\dfrac{f}{n}$

$\Rightarrow v-nf=f$

$\Rightarrow v=f+nf$

$\Rightarrow v=(n+1)f$.

But n >1

This means that (n+1) > 2.

This also means that (n+1)f > 2f

This finally means that v > 2f.

Therefore, the value of v is larger than 2f for the points above the point P.

**So, the correct answer is “Option C”.**

**Note:**

Other than taking the value of v as v =nu, we can just take some value of v greater than u, say v = 2u.

$\Rightarrow u=\dfrac{v}{2}$.

Substitute this value of u in (i).

$\Rightarrow v=\dfrac{\dfrac{v}{2}f}{\dfrac{v}{2}-f}$

$\Rightarrow \dfrac{v}{2}-f=\dfrac{f}{2}$

$\Rightarrow v-2f=f$

$\Rightarrow v=3f$

Substitute u = v in (i).

$\Rightarrow v=\dfrac{vf}{v-f}$

$\Rightarrow v-f=f$

$\Rightarrow v=2f$

This means that the value of v at point P is equal to 2f.

Hence, we can say that the value of v will be greater than 2f above point P.

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