can u plz help me out on this one..
A point source of light is placed at a certain depth below the surface of a large deep lake.show that the fraction of light energy that escapes directly from water is given by
1/2 -1/2n√n^2 -1
n being the refractive index of water....
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4 Answers
62I will give you a hint. See if you can work out the whole solution...
You have to find the angle of the conical shape out of which light can pass out.
That is given by Snell's Law.
Now once you have that angle you have to find the solid angle and divide it by the total solid angle! which is 4pi
But this second part is not in syllabus.
Both IIT JEE and CBSE.
But in any case try to find the radius of the circle that will be formed vertically above the source of light through with all light will pass out.
62I do not want to give the whole soulution becasue I think that you should try the part that is in syllabus.
That itself is an awesome solution and you will love it if you solve it yourself.
Do let us know if you cant solve this one.
1In particular, the solid angle of a symmetric cone of opening angle θ where θ is the critical angle....2pi(1-cosic )
so ..
fractional amt of light energy escaping is = 2pi(1-cosic)/4pi= 1/2(1-cosic)
again 12(1 - cosic)= 1/2[1-√(1-sin^2)]
again sin^2ic=1/μ^2=1/n^2
putting the value in the equation ....i am getting the required result.......
am i correct ....plz show if thr is any other way???