How to determine Refractive Index without using the Refractive Angle?

AI Thread Summary
The discussion focuses on determining the refractive index through a lateral displacement experiment involving a laser beam passing through a glass block in air. Participants clarify the angles involved in the experiment and emphasize the importance of consistency in the diagram presented. The conversation includes references to inverse trigonometric functions and the relationship between sine and cosine to eliminate variables in the equation for lateral displacement. Additionally, the experiment is noted for its relevance in IGCSE practical assessments, where students measure and calculate without needing to follow theoretical principles. Understanding these concepts is crucial for accurately relating lateral displacement to the incident angle.
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Homework Statement
A Lateral Displacement Experiment of a laser beam pass through a rectangle glass block in environment, AIR (n1=1.00). By measuring the Incident Angle and Lateral Displacement on the Screen WITHOUT measuring Refractive Angle. Form an equation to relate the lateral displacement (d) and incident angle (θ1).
Relevant Equations
Snell's Law
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jonas_28 said:
Homework Statement:: A Lateral Displacement Experiment of a laser beam pass through a rectangle glass block in environment, AIR (n1=1.00). By measuring the Incident Angle and Lateral Displacement on the Screen WITHOUT measuring Refractive Angle. Form an equation to relate the lateral displacement (d) and incident angle (θ1).
Relevant Equations:: Snell's Law

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View attachment 284793
Hi @jonas_28. Welcome to Physics Forums.

Your diagram shows angles ##\theta_i## and ##\theta_0##. I guess these are what you call ##\theta_1## and ##\theta_2##. Also the diagram doesn't show distance ##l## but I guess this is what is shown as AB on the diagram. It is worth checking for consistency before posting.

Are you familiar with inverse trig' functions? If ##sin(y) = x## then ##y = sin^{-1}x##.

In your working x is ##\frac {n_1}{n_2} \sin\theta_1## and y is ##\theta_2##.
 
You know that ##\sin\theta_2=\dfrac{n_1}{n_2}\sin\theta_1## and you want to eliminate ##\cos\theta_2## from your expression for ##d##. Well, you also know that ##\cos^2\theta_2+\sin^2\theta_2=1.## So ##\dots##
 
A version of this experiment appears fairly often in one of the IGCSE 'alternative to practical' papers - but as a practical exercise without the theory, ie the students measure it and perform the calculations without having to follow the theory. (eg Q5 here)

For those who don't know, IGCSE is the International version of the UK's General Cert of Secondary Education, with exams sat at age 15-16.
 
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