i) The incident rays, refracted rays and normal lie in the equivalent plane.
ii) The incident ray and refracted ray lie on opposite sides of normal.
iii) The ratio of velocity of light in rarer medium to velocity of light in denser medium is a constant called refractive index of denser medium with respect to rarer medium.
Explanation laws of refraction :
i) Let consider XY be the plane refracting surface separating two media air and glass of refractive indices μ1 and μ2 serially.
ii) A plane wavefront AB is developing at an angle towards XY from the air medium. It is bounded by the rays AA1 and BB1 which are incident rays.
iii) When point 'A' reaches 'A1' then point 'B' will be at 'P'. It even now has to cover distance PB1 to reach XY.
iv) According to Huygens' principle the secondary waves will originate from A1 and will spread over a hemisphere in the glass.
v) All the rays between AA1 and BB1 will reach XY and spread over the hemispheres of increasing radii in the glass. The surface of tangential of all such hemispheres is RB1. This allow to rise refracted wavefront B1R in the glass.
vi) A1R and B1R1, are refracted rays.
vii) Let consider c1 and c2 be the velocities of light in air and glass .
viii) At any instant of time 't' distance covered by the incident wavefront from point P to B1 = PB1 = c1t.
The distance covered by the secondary wave from point A1 to R = A1R = c2t.
Proof of laws of refraction:
i) From figure,
angle AA1M + angle MA1P = 90° .................................(1)
and angle MA1P+ angle PA1B1 = 90° ...........................(2)
From equations (1) and (2),
angle AA1M = angle PA1B1 = i
ii) Similarly, angle NA1R = angle N1B1R1 = r
We have, angle N1B1R1 + angle N1B1R = 90°.............(3)
and angle N1B1R + angle A1B1R = 90° ......................(4)
From equations (3) and (4)
angle N1B1R1 = angle A1B1R = r
iii) In triangle A 1PB1, sin i = PB1/A1B1 = c1t/A1B1................(5)
iv) In triangle A1RB1, sin r = A1R/A1B1 = c2t /A1B1..............(6)
v) Dividing equation (5) by (6),
Therefore sin i/ sin r = c1/ c2....................(7)
Also c1/ c2 = μ2/ μ1 = 1μ2 ......................(8)
Where as 1μ2 = R.I. of water with respect to air,
sin i/ sin r = μ2/ μ1
vi) From the explanation, it is clear that incident rays AA, BB, refracted rays A,R, B,R and normal MN and MIN lie on the same plane XY. Also incident ray AA, and refracted ray AR lie on opposite sides of normal MN. Therefore, laws of refraction can be explained.
THE END
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