Finding angle of refraction given angle of incidence?

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SUMMARY

The discussion focuses on calculating the angle of refraction using Snell's Law, specifically for a scenario involving a critical angle of 48 degrees at the air-water boundary. Participants clarify that the angle of incidence is 90 degrees, leading to the conclusion that the refractive index (n) can be derived from the formula n = 1/sin(C), where C is the critical angle. The calculations confirm that the refractive index is 1/sin(48), which is essential for understanding light behavior at the interface of different media.

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  • Understanding of Snell's Law and its application in optics
  • Knowledge of critical angles and refractive indices
  • Basic geometry related to angles and triangles
  • Familiarity with sine functions and trigonometric calculations
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Martin Hanna
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Can someone help me in solving this question ?

my attemp is sin42/sin 90 to get the refractive index then n =sin i/sin r to get c
and for b n=1/sinc
Physics.jpg
 
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Welcome to PF!

Try to answer part (i) first by examining the diagram and using a little geometry.
 
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Thanks for responding but i am trying but can't figure it out its probably right there in my face but i just can't see it yet
 
What is the value of the angle DBM shown below?
 

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that would be 90 ?
 
Yes, I think that's what you are supposed to assume. The fisherman's net (at D) is assumed to be essentially at the surface of the water.
 
so there fore the angle of incidence is 90 so that means C would be a critical angle
 
Martin Hanna said:
so there fore the angle of incidence is 90 so that means C would be a critical angle
Yes. From the diagram, can you determine the value of the critical angle in degrees?
 
42 ? i doubt you can just determine it from looking at the diagram as the question asks to calculate it
 
  • #10
Note that the line connecting point A to Nemo's eye is a straight line.
 

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  • #11
therefore c must be 180-(42+90)
 
  • #12
Yes
 
  • #13
well it makes sense because the answer is 48 and 48 is the critical angle of air-water boundary so now i find the refractive index 1/48
 
  • #14
Why 1/48?
 
  • #15
because there is a formula which says that N(Refractive index) = 1/ sin C(48) oh i see yeah I am suppose to sin the 48
 
  • #16
Yes, good.
 
  • #17
thank you very much
 

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