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.

**Related Posts**

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