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**Hint:**The above given problem can be solved using the speed equation in which the time of the present remains the same with respect to the velocity of the plane mirror and the speed of the image that appears on the plane mirror.

**Useful formula:**

Speed of the approaching image is given by;

$s = 2v$

Where, $s$ denotes the speed of the approaching image, $v$ denotes the speed of the approaching plane mirror.

**Complete step by step solution:**

The data given in the problem is;

The speed of the moving mirror, $v = 10\,\,m{s^{ - 1}}$.

Since the object is $x$ number of meters from the mirror and the image is at $x$ number of meters from the mirror, therefore the image distance from the mirror on the plane mirror is $2x$, that is the distance is doubled. On using the speed equation if the interval doubles, with time holds the same, then the speed would also double.

Let us assume that in a plane mirror if the mirror moves at speed of $v$, then the image that appears on the plane mirror moves at a speed of $2v$. That is in this case the mirror is moving with the speed of $v = 10\,\,m{s^{ - 1}}$ .

That is the speed of the image that is moving and approaching us is;

$s = 2v$

Substitute the values of the speed by which the plane mirror that is approaching towards us;

$

s = 2 \times 10\,\,m{s^{ - 1}} \\

s = 20\,\,m{s^{ - 1}} \\

$

Therefore, the speed by which the image will approach us is $s = 20\,\,m{s^{ - 1}}$

**Hence, the option (D), $s = 20\,\,m{s^{ - 1}}$ is the correct answer.**

**Note:**The mirror formula is related only to the plane mirror. In the plane mirror formula, the focal length value is zero that is the principle focus is at infinity. Mirror formula expresses that the connection incorporates the image distance, the object distance and the focal length of the mirror.

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