Question about light reflection

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SUMMARY

The discussion focuses on calculating the angle at which light exits water when a diver shines a flashlight upward at a 23.0° angle to the vertical. The correct approach involves applying Snell's Law, which states that n1 * sin(θ1) = n2 * sin(θ2). Here, n1 is the index of refraction of water (1.333), θ1 is the angle of incidence (23.0°), and n2 is the index of refraction of air (1). The calculation leads to determining the angle of refraction (θ2) as the light exits the water.

PREREQUISITES
  • Understanding of Snell's Law
  • Knowledge of the index of refraction (n) values for water and air
  • Basic trigonometry, specifically sine functions
  • Ability to manipulate equations to solve for unknown variables
NEXT STEPS
  • Study Snell's Law in detail, including derivations and applications
  • Learn about the index of refraction for various materials
  • Practice problems involving light refraction at different angles
  • Explore real-world applications of light refraction in optics
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Students studying physics, particularly those focusing on optics and light behavior, as well as educators looking for examples of Snell's Law applications.

kaptinmorgan
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Homework Statement



A diver shines a flashlight upward from beneath the water at a 23.0° angle to the vertical. At what angle does the light leave the water?

Homework Equations


I thought that i could use snell's law here...but i don't know what to do afterwards...



The Attempt at a Solution



angle of incidence is 23... and the index of refraction is 1.333

1.333 (sin 23) = 1 (sin x)

is that about right?
 
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kaptinmorgan said:
angle of incidence is 23... and the index of refraction is 1.333

1.333 (sin 23) = 1 (sin x)

is that about right?

Write very clearly what x represents here. What do you feel that the question wants?
 

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