Question

A crow is flying at a height 20m above the water surface. A fish is located directly below the crow at a dept 10m below the surface of water .If velocity of crow towards fish is ${{v}_{c}}=6m{{s}^{-1}}$and velocity of fish towards crow is ${{v}_{f}}=2m{{s}^{-1}}$then velocity of crow seen by fish in a given refractive index of water$\mu =\dfrac{4}{3}$?

& A)5m{{s}^{-1}} \\\ & B)10m{{s}^{-1}} \\\ & C)15m{{s}^{-1}} \\\ & D)20m{{s}^{-1}} \\\ \end{aligned}$$

Answer 1

Here in this type of question when any object is lying below water then with respect to that object image of object outside the water is slightly raised due to refraction of light phenomenon .So here image of crow with respect to fish is slightly above the original position of crow and this raises image depends on the refractive index of given liquid.

Complete step by step answer:
Here in the figure it is shown that the crow is above 20m above the water surface and fish is below 10 m directly below the water surface.

Here fish is below water so due to the refraction of light from water to air, the image of crow is slightly displaced up from its original position. Like in the case of a coin placed in a glass full of water it is seen that it is slightly above from its original position. Similarly crow is slightly above its original position. This new distance of crow with respect to fish is dependent on the refractive index of liquid.
Here, Let us assume that the distance of fish above the water surface is y Where y = 10 m Let us assume the distance of the crow below the water surface is x. Where x= 10 m Image of crow seen by fish at some distance above the water surface. So this Distance of crow seen by fish is represented by D =$y+\mu x$.
Now we will calculate the relative velocity of the crow with respect to fish. This can be calculated by differentiating the distance equation with respect to time. Now differentiating this distance equation we get, Let us assume the velocity of crow with respect to fish is given by V. Velocity of crow with respect to fish = $\dfrac{dD}{dt}$ $V=\dfrac{dD}{dt}$ $\Rightarrow V=\dfrac{d}{dt}(y)+\mu \dfrac{d}{dt}(x)$ It is given in the question that the velocity of fish towards crow is ${{v}_{f}}=2m{{s}^{-1}}$. Velocity of crow towards fish is ${{v}_{c}}=6m{{s}^{-1}}$ $\Rightarrow V={{v}_{f}}+\mu \times {{v}_{c}}$ Now these value of velocities in above equation we get ,

& \Rightarrow V=2+\dfrac{4}{3}\times 6 \\\ & \therefore V=10m{{s}^{-1}} \\\ \end{aligned}$$ **So, the correct answer is “Option B”.** **Note:** Refractive index plays an important role in the phenomenon of refraction because it gives us the information about the bending of light when it passes from one medium to another. Dependence of Refractive index of following factors:- (i) Nature of the media of incidence and refraction. (ii) Colour of light or wavelength of light. (iii) Refractive index decreases with the increase in temperature.