THE PRACTICE STUDIO
Go slowly. Ask why. Make it yours.
PRACTICE / Snell law
A ray crosses into glass
A ray in air with index 1 enters glass of index 1.5 at incidence angle 30° from the normal.
Find sin r for the refracted ray.
Organize the givens & assumptions
- n₁ = 1
- n₂ = 1.5
- i = 30°
Smooth plane interface between isotropic media; angles measured from the normal; indices given at the light frequency.
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Opening this prompt marks this practice attempt as assisted.
Problem: A ray crosses into glass A ray in air with index 1 enters glass of index 1.5 at incidence angle 30° from the normal. Question: Find sin r for the refracted ray. Givens: - n₁ = 1 - n₂ = 1.5 - i = 30° Assumptions: Smooth plane interface between isotropic media; angles measured from the normal; indices given at the light frequency. Choices (LaTeX): A. 3/4 B. 1/2 C. 2/3 D. 1/3 Ask for my approach first. Help me distinguish a physical misconception from an algebra error without revealing which choice is correct. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
Start with the smallest hint.
Work through it with me.
One decision at a time. Pause, predict, then unpack the algebra.
The larger refractive index makes the angle smaller, so its sine must be below one half.
Choose a physical relation
Continuity of phase along the interface links the ray angles through the refractive indices.
Isolate what is asked
The larger refractive index makes the angle smaller, so its sine must be below one half.
Why is this step valid?
Divide by n₂ and take inverse sine; sin r = 1/3 is not itself the angle in degrees.
Substitute and interpret
Refraction angle is about 19.5°, smaller than incidence because the ray enters the higher-index medium.
The proposed sine satisfies the interface phase-matching condition in Snell’s law.
TRY THE IDEA AGAIN
Can total internal reflection occur from air into this higher-index glass?
YOUR LEARNING NOTE
How does this idea feel now?
Confidence is a reflection, not proof of mastery.
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