Snell's law, critical angle, and angle of incidence

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Homework Help Overview

The discussion revolves around the application of Snell's law in determining the critical angle of refraction when light passes from acetone into air. Participants are exploring the relationship between the angle of incidence and the angle of refraction, particularly in the context of critical angles.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the critical angle and question the wording of the problem regarding the refracted angle. There is confusion about whether the critical angle corresponds to the angle of incidence or the angle of refraction.

Discussion Status

Some participants have provided calculations and interpretations of the critical angle, while others are questioning the assumptions made about the indices of refraction. There is an ongoing exploration of the implications of the problem's wording and the values used in calculations.

Contextual Notes

There are indications that the indices of refraction used may not align with standard values, leading to discrepancies in the calculations. Participants are encouraged to clarify these assumptions and the context of the problem.

reb659
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Homework Statement



A pair of students measure the refraction of light in passing from acetone into air for several angles. When the angle of incidence is 30, the angle of refraction is 42.

b) What is the angle of incidence (in degrees) when the refracted angle is the critical angle?

Homework Equations


Snells law: Ni*sin(i)=Nr*sin(r)

The Attempt at a Solution



I found the index of refraction for air to be 1 and for acetone to be 1.36. By snell's law setting r=90:
sin(i crit) = nr/ni * sin(90) = nr/ni = 1/1.36

icrit = arcsin(1/1.36) = 55.73 degrees

Is my answer correct? I'm a bit confused about the part that says "...when the refracted angle is the critical angle". Isn't the angle of incidence (when it creates a refraction of 90 degrees) the critical angle?
 
Last edited:
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I see what you mean, indeed it's worded poorly...

although realistically it would be a stupid question asking you to work out the angle of incidence where the angle of refraction is the critical angle...i would go with your workings and ask your teacher/whatever for clarification
 
Yes, it is. You went only half-way answering part (b) which is to find the critical angle. Now answer the question. You have a refracted angle of 55.73 degrees, what is the angle of incidence?
 
Why wouldn't it just be 55 degrees?
 
reb659 said:
Why wouldn't it just be 55 degrees?
-
55.73 is the angle of refraction in the air. For that what should be the angle of incidence in acetone?
 
I thought that in calculating the critical angle to be 55.73 it would be equal to the angle of incidence, not the refracted angle. Isn't the refracted angle just 90?
 
It appears in the problem that the indexes of refraction used are not the standard values you looked up. (I get a lower ratio)

From the original angles you can use Snell's law to calculate the ratio of the indexes of refraction. Then as explained find the incident angle which refracts to 90deg and that is the critical angle.

The problem is worded OK in my opinion but instead of "acetone" the problem should have stated "some unknown liquid".

EDIT: Hint: I get an angle of _ _ . _ _ _ _ 1711717478...
And it is significantly less than 55.73
 

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