Light passing from water to air is incident on the interface at an anle 1 from t

Question

Light passing from water to air is incident on the interface at an anle 1 from the normal.

Part A - Which expression gives the angle at which the light emerges into the air relative to the surface?

nwater = 4/3

nwater = 4/3

  =sin1(43sin1)

=sin1(34sin1)

=90sin1(43sin1)

Part B - Which expression gives the incident angle from the normal for which no light crosses the interface (critical angle)?

1=90sin1(34)

1=sin1(43)

  1=90sin1(43)

  

What happens to the light incident to the normal of the interface at an angle greater than the critical angle?

  =90sin1(34sin1)  

  =sin1(43sin1)

=sin1(34sin1)

=90sin1(43sin1)

Part B - Which expression gives the incident angle from the normal for which no light crosses the interface (critical angle)?

1=90sin1(34)

1=sin1(43)

1=sin1(34)   

  1=90sin1(43)

  

What happens to the light incident to the normal of the interface at an angle greater than the critical angle?

It is reflected at an angle equal to the incident angle. It is cancelled out through destructive interference. It is reflected perpendicular to the interface. It propogates along the surface of the interface.

Explanation / Answer

PART A -
Using Snell's Law -
n1 * sin(1) = n2 * sin(r)

Where,
n1 = Reractive index of water = 4/3
n1 = Reractive index of air = 1

4/3 * sin(1) = 1 * sin(r)
r = sin^-1 (4/3 * sin(1))
Expression of the angle at which the light emerges into the air relative to the surface r = sin^-1 (4/3 * sin(1))

PART B-
For critical angle , Angle of refraction = 90o
Therefore,
Using Snell's Law -
n1 * sin(1) = n2 * sin(r)
4/3 * sin(1) = 1 * sin(90)
sin(1) = 3/4
1 = sin^-1(3/4)
Expression which gives the incident angle from the normal for which no light crosses the interface,
1 = sin^-1(3/4)

Q. What happens to the light incident to the normal of the interface at an angle greater than the critical angle?
Answer - It is reflected at an angle equal to the incident angle.

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