Object O is moving with a velocity $$8\ m.s^{-1}$$ inside water and mirror is moving with a velocity $$6\ m.s^{-1}$$ of as shown in the figure. Then find the velocity of image formed by the plane mirror with respect to object in $$ (m.s^{-1}$$ )

The object is in water while the mirror is in air. Therefore, first convert the velocities into the same medium using the relative refractive index.

For water with respect to air,  $$\mu_{rel}=\frac{\mu_o}{\mu_{observer}}=\frac{4/3}{1}=\frac{4}{3}$$

Hence, the apparent depth is  $$d'=\frac{d}{\mu_{rel}}$$

Therefore, the apparent velocity of the object as seen from air is  $$v_o'=\frac{8}{4/3}=6\,m\,s^{-1}$$

Equivalently, when we consider the mirror from the water side, its apparent velocity is obtained using the inverse relative refractive index: $$\mu_{rel}=\frac{1}{4/3}=\frac{3}{4}$$

Thus,  $$v_m'=\frac{6}{3/4}=8\,m\,s^{-1}$$

Now both the object and mirror are considered in the same medium, and both have velocity  $$v_o=v_m'=8\,m\,s^{-1}$$

For a plane mirror,  $$v_i=2v_m-v_o$$

Therefore,  $$v_i=2(8)-8$$

$$v_i=8\,m\,s^{-1}$$

Hence, the image and the object have the same velocity.

Therefore, the velocity of image with respect to object is  $$v_{i/o}=v_i-v_o$$

$$v_{i/o}=8-8=0$$

Thus,  $${0\,m\,s^{-1}}$$

Hence, the correct answer is 0.