Reflection/Refraction homework help

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SUMMARY

The discussion focuses on the behavior of light rays as they pass through a glass prism with a refractive index (n) of 1.65, specifically under two different immersion conditions: air and water. For part A, when the prism is immersed in air, the light ray emerges at an angle of 50.14 degrees after passing through the prism. In part B, when the prism is immersed in water (with n approximately 1.3), the critical angle is calculated, indicating a change in the light's path due to the different refractive indices.

PREREQUISITES
  • Understanding of Snell's Law (n_a sin(theta_a) = n_b sin(theta_b)
  • Knowledge of critical angle calculations (sin(theta)critical = n_b/n_a)
  • Familiarity with refractive indices of common substances (e.g., air, water, glass)
  • Basic principles of optics and light refraction
NEXT STEPS
  • Research the concept of total internal reflection and its applications
  • Learn about the refractive indices of various materials and their impact on light behavior
  • Explore advanced optics topics such as lens design and optical systems
  • Investigate the effects of temperature and wavelength on refractive indices
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Students studying optics, physics educators, and anyone interested in understanding light behavior in different mediums.

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Homework Statement



A light ray is incident on a glass prism with n=1.65. (A) If the prism is immersed in air, where does the light ray emerge into the air? At what angle? (B) If the prism is immersed in water, where does the light emerge into the water? At what angle?

_____ A_____C
_____ |90 45/
_____ |____/
----->|___/
_____ |n /
_____ B /

Homework Equations


n_a sin(theta_a) = n_b sin(theta_b)
sin(theta)critical= n_b/n_a

The Attempt at a Solution



PART A:

AB)
1.0 sin(0)= 1.65 sin(theta)'
(theta)'= 0 (no bending)

BC) (crtitical angle 41.1 in air?)
1.65 sin 45 = 1.52 sin (theta)'
(theta)'= 50.14 <- final answer

PART B:
just change the n's?

Any good?
 
Last edited:
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B, yes n for water is around 1.3
 

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