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monochromatic light passes in three layers. Each medium are n1=2, n2=1,5, n3=1,2. Let...
monochromatic light passes in three layers. Each medium are n1=2, n2=1,5, n3=1,2. Let the frequency and the speed of light in the vacuum be 5x10^14 Hz and 3x10^8 m/s.
1. calculate the speed of light in the second layer
2. calculate the wavelenght of light in the second layer
3. if the angle of incidence `theta` at the boundary B is larger than the critical angle `theta`c the total reflection occurs. Find the value if sin `theta`c
1 Answer | add yours
The question has too many parts, only parts 1 and b) are related. Hence these two are answered.
We know, `v_0=nulambda_0`
In vacuum, `3*10^8=5*10^14*lambda_0`
`rArr lambda_0=6*10^-7` m
Again, refractive index, `n_1` , in a medium with respect to vacuum can be expressed as
`n_1 = v_0/v_1=lambda_0/lambda_1`
`n_2 = v_0/v_2=lambda_0/lambda_2`
a) Here, the second layer has a refractive index of 1.5
Therefore, `v_2=3*10^8/1.5 =2*10^8` m/s
b) Similarly, `lambda_2=6*10^-7/1.5=4*10^-7` m
Posted by llltkl on July 2, 2013 at 10:06 AM (Answer #1)
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