How far is the ray displaced as a result of traveling through the gla

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A ray of light traveling from air into dense flint glass at an angle of incidence of 63.5° experiences a displacement due to refraction. The thickness of the glass is 4.26 mm, and the calculated displacement is 5.671 mm. The calculations involve using Snell's Law to determine the angle of refraction and applying trigonometric functions to find the displacement. The final displacement value represents the distance between the actual exit point of the ray and where it would have exited if the glass had an index of refraction of 1.

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1. A ray of light passes from air through dense flint glass and then back into air. The angle of incidence on the first glass surface is 63.5°. The thickness of the glass is 4.26 mm; its front and back surfaces are parallel. How far is the ray displaced as a result of traveling through the glass?


this is what I tried...

1sin65.5=1.6sinA
sinA=0.5593
A=34.01°

x=4.26tan63.5°-4.26tan34.01°
= 5.671mm


I am no sure where i did wrong...
 
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lizaliiu said:
1. A ray of light passes from air through dense flint glass and then back into air. The angle of incidence on the first glass surface is 63.5°. The thickness of the glass is 4.26 mm; its front and back surfaces are parallel. How far is the ray displaced as a result of traveling through the glass?

this is what I tried...

1sin65.5=1.6sinA
sinA=0.5593
A=34.01°

x=4.26tan63.5°-4.26tan34.01°
= 5.671mm

I am no sure where i did wrong...
You really need to provide a sketch of the situation for us to be of much help.
 


lizaliiu said:
x=4.26tan63.5°-4.26tan34.01°
= 5.671mm
That's the distance between the point where the ray exited the glass and the point where it would have exited the glass had its index been 1. That is not the same as the distance between the two paths.
 

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