Snell's law physics question help needed!

[dorjiako]

Help needed in solving this question.
A piece of gallium phosphide is frozen in ice. A beam of light is directed downward through the ice-gallium phosphide boundary at a 25 degrees angle from the normal. The light emerges from the gallium phosphide 12.25mm away from where it would have had the solid been pure ice. Find the thickness of the gallium phosphide layer. Gallium phosphide has the highest known optical density (3.5), and ice has the third lowest (1.31).

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[BerkReviewTeach]

Help needed in solving this question.
A piece of gallium phosphide is frozen in ice. A beam of light is directed downward through the ice-gallium phosphide boundary at a 25 degrees angle from the normal. The light emerges from the gallium phosphide 12.25mm away from where it would have had the solid been pure ice. Find the thickness of the gallium phosphide layer. Gallium phosphide has the highest known optical density (3.5), and ice has the third lowest (1.31).

Seriously? That is a geometry question that sounds more like a homework question in school than anything even close to what the MCAT would ask. If you are trying to get homework answered, then you might want to at least give four answer choices.

a) d = arctan [(sin 25degrees)/(1.31/3.5)]degrees x 12.25
b) d = arctan [(sin 25degrees)/(1.31/3.5)]degrees / 12.25
c) d = 12.25/tan [arcsine (1.31/3.5 x sin 25degrees)]degrees
d) d = 12.25/tan [arcsine (3.5/1.31 x sin 25degrees)]degrees

To solve it, you'll need to draw a right triangle involving the exiting surface of gallium phosphide on bottom, the refacted ray as one side, and the normal extending from the first interface to the second interface as the remaining side (which is the length you're solving for). You know the angle of refraction from Snell's law and they gave you the length of the bottom segment of that triangle. From there it's either a creative use of Pythagorean theorem or trigonometry invloving the tangent (opp/adj) of the refacted angle. Good luck!

 

[dorjiako]

Seriously? That is a geometry question that sounds more like a homework question in school than anything even close to what the MCAT would ask. If you are trying to get homework answered, then you might want to at least give four answer choices.

a) d = arctan [(sin 25degrees)/(1.31/3.5)]degrees x 12.25
b) d = arctan [(sin 25degrees)/(1.31/3.5)]degrees / 12.25
c) d = 12.25/tan [arcsine (1.31/3.5 x sin 25degrees)]degrees
d) d = 12.25/tan [arcsine (3.5/1.31 x sin 25degrees)]degrees

To solve it, you'll need to draw a right triangle involving the exiting surface of gallium phosphide on bottom, the refacted ray as one side, and the normal extending from the first interface to the second interface as the remaining side (which is the length you're solving for). You know the angle of refraction from Snell's law and they gave you the length of the bottom segment of that triangle. From there it's either a creative use of Pythagorean theorem or trigonometry invloving the tangent (opp/adj) of the refacted angle. Good luck!

I will try what you said and see if I can get their answer 42.2mm. I was just having problem in my perception of the question. I did not know if I should assume that the light was from air to ice-gallium boundary and out to water or .....

 

[dorjiako]

The diagram you explained above is not working out the answer. Is it possible that there are certain steps I am missing. Help is still needed!

Using your explained diagram, the refracted angle (assuming the light is going from ice to gallium) will be (1.31sin25degrees/3.5) = 9.1 degrees.
Tan 9.1 degree = 12.25/d
d= 12.25/0.1602 = 76.5m .However, the answer is 42.2mM.
Hence, this came down to my initial confusion, how exactly is the light travelling (is is from air to ice-gallium and out to air .....) and what exactly does 12.25mm represent in this question.

Help still needed!

 

[FutureDoctor503]

The diagram you explained above is not working out the answer. Is it possible that there are certain steps I am missing. Help is still needed!

Using your explained diagram, the refracted angle (assuming the light is going from ice to gallium) will be (1.31sin25degrees/3.5) = 9.1 degrees.
Tan 9.1 degree = 12.25/d
d= 12.25/0.1602 = 76.5m .However, the answer is 42.2mM.
Hence, this came down to my initial confusion, how exactly is the light travelling (is is from air to ice-gallium and out to air .....) and what exactly does 12.25mm represent in this question.

Help still needed!

Your answer "9.1 degrees" is wrong. Check your calculations.

 

[dorjiako]

Your answer "9.1 degrees" is wrong. Check your calculations.

I just did the calculation again and it is 9.1 degrees.
Here it is= 1.31sin25/3.5= 0.158
sin^-1 of 0.158 = 9.1 degrees.
Let me know where I made the mistake so that I can move on from this question that has consumed my whole time today!

 

[FutureDoctor503]

I just did the calculation again and it is 9.1 degrees.
Here it is= 1.31sin25/3.5= 0.158
sin^-1 of 0.158 = 9.1 degrees.
Let me know where I made the mistake so that I can move on from this question that has consumed my whole time today!

I did it on google calculator multiple times and this is what I am getting http://www.google.com/#sclient=psy&...gc.r_pw.&fp=82f9f46bc5e7e168&biw=1280&bih=549

When I tried it using the calculator on the computer, I got the same answer you did. What calculator are you using?

 

[dorjiako]

I did it on google calculator multiple times and this is what I am getting http://www.google.com/#sclient=psy&...gc.r_pw.&fp=82f9f46bc5e7e168&biw=1280&bih=549

When I tried it using the calculator on the computer, I got the same answer you did. What calculator are you using?

I truly need help help. Your answer was different from mine because your calculator was in radian mode.

 

[FutureDoctor503]

On google, the answer is in radians. So, you did get the right answer.

http://ca.answers.yahoo.com/question/index?qid=20100603215106AAepROP

I tried to do the problem using this method.

 

[dorjiako]

since we are geting different answer from the book, then the answer 42.2mm is probably not right!!

 

[FutureDoctor503]

This is from another forum. It's the same question.

http://www.physicsforums.com/showthread.php?t=405877

 

[dorjiako]

This is from another forum. It's the same question.

http://www.physicsforums.com/showthread.php?t=405877

Although, I have spent ridiculous amount of hours today in figuring out this question. I will now move on to another practice question because you provided me the link that has finally satisfied my quest. Thanks again.