Refraction at Plane Surface
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Refraction
Rarer Medium and Denser Medium
Laws of Refraction of Light
i. The incident ray, refracted ray and the normal at the point of incidence lie in the same plane.
ii. When light travels from a rarer medium to a denser medium, it bends towards the normal. When light travels from a denser medium to a rarer medium, it bends away from the normal.
iii. For a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant. This constant is called the refractive index. This is Snell’s law.
The notation used in the scan is of the form aμb, meaning the refractive index of medium b with respect to medium a, or equivalently light travelling from medium a to medium b.
Refractive Index
where:
Principle of Reversibility of Light
Consider a ray travelling from air to water. Let i be the angle of incidence and r the angle of refraction.
Refractive index of water with respect to air:
aμw = sin i / sin r … (i)
When the ray is reversed and travels from water to air:
wμa = sin r / sin i … (ii)
Multiplying (i) and (ii):
aμw · wμa = 1
wμa = 1 / aμw
Real Depth and Apparent Depth
When a light ray travels from a denser medium to a rarer medium, it bends away from the normal. Due to this, the bottom of a pond appears to be raised.
Let:
For refraction from water to air, the source uses:
aμw = sin r / sin i
From the triangles in the figure:
sin r = CA/O′A
sin i = CA/OA
Therefore:
aμw = OA/O′A
If point A is very close to C, OA ≈ OC and O′A ≈ O′C:
aμw = Real depth / Apparent depth
Hence:
Apparent depth = Real depth / aμw
Apparent Shift
If the real depth is t:
Apparent depth = t / aμw
d = t − t/aμw
d = t[1 − 1/aμw]
Lateral Shift
Consider a glass slab of thickness t. A ray is incident at angle i, refracted inside the slab at angle r, and emerges from the lower face. Let the lateral shift be d.
From the triangle inside the slab:
cos r = t/OB
Therefore:
OB = t/cos r … (i)
From the triangle used for lateral displacement:
sin(i − r) = d/OB
d = OB sin(i − r)
Using (i):
d = t sin(i − r) / cos r
Special Case: i = 90°
d = t sin(90° − r)/cos r
sin(90° − r) = cos r
d = t
The source concludes that when the angle of incidence is 90°, the lateral shift equals the thickness of the glass slab.
Total Internal Reflection and Critical Angle
Conditions for Total Internal Reflection
1. The source states that the object/light must be in the denser medium, so that light travels from denser medium toward rarer medium.
2. The angle of incidence must be greater than the critical angle: i > C.
When light travels from a denser medium to a rarer medium, it bends away from the normal. As the angle of incidence is increased, the angle of refraction also increases.
When the angle of incidence is increased beyond the critical angle, the ray returns into the same denser medium. This phenomenon is called total internal reflection.
Relation Between Refractive Index (μ) and Critical Angle (C)
The source considers a glass–air boundary. A ray in glass is incident at the critical angle C and is refracted along the surface, so r = 90°.
For air with respect to glass, the source writes:
gμa = sin i / sin r
At the critical angle:
i = C and r = 90°
Therefore:
gμa = sin C / sin 90°
gμa = sin C … (i)
From the reversibility relation:
aμg = 1 / gμa
Using (i):
aμg = 1 / sin C
This is the relation between the refractive index of the denser medium with respect to the rarer medium and its critical angle, as presented in the source notes.
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