What Determines the Critical Angle in a Glass-to-Liquid Interface?

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SUMMARY

The critical angle at a glass-to-liquid interface is determined by the indices of refraction of the two media. In this case, the refractive index of glass is 1.52 and that of the liquid is 1.75. Since the light travels from a denser medium (glass) to a less dense medium (liquid), there is no critical angle, as confirmed by the equation critical angle = inverse sin (n2/n1). Attempting to calculate the critical angle using inverse sin (1.52/1.75) yields an angle of 60.3 degrees, but this is incorrect as it does not meet the conditions for total internal reflection.

PREREQUISITES
  • Understanding of Snell's Law (n1 sin i1 = n2 sin i2)
  • Knowledge of refractive indices and their implications
  • Familiarity with the concept of total internal reflection
  • Basic trigonometry, specifically the inverse sine function
NEXT STEPS
  • Study the principles of total internal reflection in optics
  • Learn about the calculation of critical angles in various media
  • Explore the applications of Snell's Law in real-world scenarios
  • Investigate the effects of varying refractive indices on light behavior
USEFUL FOR

Students studying optics, physics educators, and anyone interested in the principles of light behavior at interfaces between different media.

sgoeke
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Homework Statement


A ray of light travels across a glass-to-liqued interface. If the indices of refraction for the liquid and glass are, respictevly 1.75 and 1.52, what is the critical angle at this interface?


Homework Equations


critical angle = inverse sin (n2/n1)


The Attempt at a Solution


i used inverse sin (1.52/1.75) and got an angle of 60.3 degrees. However, I do know that if you reverse the 1.75 and 1.52 in that equation, you get an undefined angle and the correct answer is that there is not critical angle. Why do you used 1.75/1.52?
 
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sgoeke said:

Homework Statement


A ray of light travels across a glass-to-liqued interface. If the indices of refraction for the liquid and glass are, respictevly 1.75 and 1.52, what is the critical angle at this interface?


Homework Equations


critical angle = inverse sin (n2/n1)


The Attempt at a Solution


i used inverse sin (1.52/1.75) and got an angle of 60.3 degrees. However, I do know that if you reverse the 1.75 and 1.52 in that equation, you get an undefined angle and the correct answer is that there is not critical angle. Why do you used 1.75/1.52?

You get a critical angle and total internal reflection when the light goes from a dense to a less dense medium. That is, one with a refractive index that is higher to one that is lower.
The question says that the light goes from glass to liquid and that glass has index=1.52 and the liquid 1.75.
So there is no critical angle in this case.

The law is n1 sin i1 = n2 sin i2

For the critical angle, i2 = 90 so, sin i2 = 1
 

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