An object is moving towards a stationary plane mirror with a speed of \[2m{s^{ - 1}}\]in left. Velocity of the image w.r.t. the object is:
a. \[2m{s^{ - 1}}\]towards right
b. \[4m{s^{ - 1}}\]towards right
c. \[2m{s^{ - 1}}\]towards left
D. \[4m{s^{ - 1}}\]towards left

Hint: When it comes to solving this type of question from optical physics we must know the concepts clearly that whenever there is any object placed in front of the mirror and it’s moving with any speed then the reflection (image) also moves with same speed in the opposite direction of the movement of the object.

Complete answer:
When the object is kept in front of the mirror which is stationary and plane and that object is moving with the speed of \[2m{s^{ - 1}}\]in left of the mirror.

For this we have to draw a diagram that will explain the situation presented in the question.

Here, in the above figure we get to know that the object and image are identical and is known as the first law of reflection, size and shape remains same there will be no change. Second thing is that the speed of will remain same in magnitude but different in the direction.

So, let us say \[{v_o}\] be the velocity of the object and \[{v_i}\] be the velocity of the image. Thus according to given condition the left of the mirror represents similarity to the number line that means left side will be positive numbered side and right side will be negative numbered side.

Thus the direction being opposite the velocities are given by:
\[{v_o} = - 2m{s^{ - 1}}\] and \[{v_i} = 2m{s^{ - 1}}\]

Negative because moving in opposite directions.
Thus the speed of the with respect to the object \[ = {v_i} - {v_o}\]
\[ = 2m{s^{ - 1}} - ( - 2m{s^{ - 1}})\]
\[ = 4m{s^{ - 1}}\]

So, the answer is \[4m{s^{ - 1}}\]towards left of the mirror. Correct option is d.

Note: Here, the image velocity is to be considered as positive since its already representing negative number and with addition to the direction it becomes positive that is why we wrote here in positive format.