## Saturday, January 24, 2015

### Problems on normal shift and Refraction of Light

Normal shift happens when the object and the corresponding image are in two different mediums. Depending on the location of the object with respect to the medium there will be the normal shift. For example if the object is in the denser medium and observer is in the rarer medium, for the observer the object appears nearer to the surface. If the object is in the rarer medium and observer is in the denser medium, for the observer the object appears far from the original location.

Basing on the derivation is that we have made in the previous posts, we can use the equations that are derived for the normal shifts in the possible cases.

Problem and solution

A layer of oil 3 cm thick is floating on the layer of color water 5 cm thick. The refractive index of color water is 5 / 3. If the apparent depth of the two liquids together reuse 36 / 7 cm,  what is the refractive index of oil ?

To solve this problem let us imagine a small colliery at the bottom of the vessel, below the two liquids. Let you yourself is the observer who is observing the oil from the surface. Therefore here in this case the object is in the denser medium and observer is in the rarer medium. As a result the object appears near to the surface. We can define refractive index in this case as the ratio of real depth to the apparent depth.

By applying this formula to both the liquids together we can get the refractive index of the given liquid as shown below.

Problem and solution

A fish is rising vertically to the surface of water in LA uniformly at a rate of 3 m/s. It observes a bird diving vertically towards the water at the rate of 9 m/s above it. If the refractive index of water is 4 /3, find the actual speed of the bird above the fish?

In this problem the fish is under the water and the bird in the air. Bird is the object and fishes the observer. As the object is the rarer medium and observer is the denser medium, for the observer objects appears away from its original position. Therefore the totally distance between the fish and the bird is sum of depth of the fish from the surface and the apparent height of the bird from the surface.

As the problem is dealing with the velocity, to get the velocity from the displacements we have to differentiate displacement once with respect to time.

We need to differentiate because rate of change of displacement is defined as velocity. In this problem the relative velocity of the bird with respect to fish is  9 m/s. We have to calculate the velocity of the bird with respect to air.

By simplifying the above equation we can get the velocity of the bird as shown.

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