The 45-45-90 triangular prism shown in the figure has an index of refraction of 1.44, and is surrounded by air. A ray of light is incident on the left face at an angle of 46.0°. The point of incidence is high enough that the refracted ray hits the opposite sloping side.
Calculate the angle at which the ray exits the prism, relative to the normal of that surface. Express your answer as a negative number if the angle is clockwise relative to the normal, and a positive number for a counterclockwise angle.
I have tried this problem over and over and I don't know if it's an input error or my work is off so please look at this work and tell me what am I doing wrong. Thanks!
In order to answer this part, i know I need the incidence ray on the bottom surface and this is the steps i took to get there:
The refracted angle is arcsin(sin(46.0)/1.44) = 31.94. Thus the angle at the left interface in the upper triangle is 90 - 31.94 = 58.06
Therefore on the right side the angle in the upper triangle is 180 - 90- 58.06 = 31.94 the incident angle on the right face is 58.06....
the critical angle for this prism is arcsin(1/1.44) = 48.87
So this ray is reflected internally. so the reflected angle is 54.33o....
So the angle in the right hand triangle is 35.67o...So this reflected ray strikes the lower surface at an angle with respect to the bottom of 180 - 48.86- 31.94 = 99.2
This means the incident ray on the bottom surface is 9.20
After I got did that I calculated: arcsin(1.44*sin(9.20)) = 13.30
It keeps telling me Im wrong. Here are my submissions: 14.90 deg, 13.30 deg and 13.28 deg
Second part of question: What is the maximum index of refraction for which the ray exits the prism from the right-hand side? (It may help to remember that sin(90° - θ) = cos θ.)
Heres my work: For the ray to exist the right hand side we have from the left hand side
sin(46.0) = n*sin(θ) and at the rt hand side the incident angle will be (90 - θ)
so n*sin(90 - θ) = 1 but sin(90 - θ) = cos(θ) so n*cos(θ) = 1
So if we square both eqns we get
sin^2(46.) = n^2*sin^2(θ)
and 1 = n^2*cos^2(θ)...
Now adding these we get
1+sin^2(46.0) = n^2*(sin^2(θ) + cos^2(θ)) = n^2
Therefore n = sqrt((1+sin^2(46.0))= 1.288 = 1.29
Here are my submissions:
Incorrect 1.29
2 Incorrect. (Try 2) 1.31
3 Incorrect. (Try 3) 1.28
4 Incorrect. (Try 4) 1.26
5 Incorrect. (Try 5) 1.27
Please help! What am I doing wrong?
To calculate the angle at which the ray exits the prism, relative to the normal of that surface, we need to follow these steps:
1. Calculate the angle of incidence on the right face:
- The angle of incidence on the left face is given as 46.0°.
- Since the prism has an index of refraction of 1.44, we can use Snell's Law to find the angle of refraction on the left face. The formula is sin(theta1) / sin(theta2) = n2 / n1, where theta1 is the angle of incidence, theta2 is the angle of refraction, and n1 and n2 are the indices of refraction of the initial and final medium respectively.
- Plugging in the values, we have sin(46.0°) / sin(theta2) = 1.44 / 1.00 (since air has an index of refraction close to 1.00).
- Solve for sin(theta2): sin(theta2) = sin(46.0°) / 1.44.
- Calculate the angle of refraction on the left face: theta2 = arcsin(sin(46.0°) / 1.44).
- The angle at the left interface in the upper triangle is 90° - theta2 = 90° - arcsin(sin(46.0°) / 1.44).
2. Calculate the angle of incidence on the right face:
- The angle in the upper triangle on the right side is the same as the angle on the left side, since the prism is symmetrical.
- Therefore, the angle of incidence on the right face is also 90° - arcsin(sin(46.0°) / 1.44).
3. Calculate the angle of reflection on the right face:
- Since the angle of incidence on the right face is greater than the critical angle (48.87°), total internal reflection occurs.
- The reflected angle is equal to the angle of incidence, so the angle of reflection on the right face is 90° - arcsin(sin(46.0°) / 1.44).
4. Calculate the angle at which the ray strikes the bottom surface:
- The angle in the right-hand triangle is the same as the angle of reflection on the right face.
- Therefore, the incident angle on the bottom surface is 180° - (90° - arcsin(sin(46.0°) / 1.44)) - (90° - arcsin(sin(46.0°) / 1.44)).
5. Calculate the angle of refraction on the bottom surface:
- Apply Snell's Law again. The formula is sin(theta1) / sin(theta2) = n2 / n1, where theta1 is the angle of incidence, theta2 is the angle of refraction, and n1 and n2 are the indices of refraction of the initial and final medium respectively.
- Plugging in the values, we have sin(180° - (90° - arcsin(sin(46.0°) / 1.44)) - (90° - arcsin(sin(46.0°) / 1.44))) / sin(theta2) = 1.44 / 1.00 (since air has an index of refraction close to 1.00).
- Solve for sin(theta2): sin(theta2) = sin(180° - (90° - arcsin(sin(46.0°) / 1.44)) - (90° - arcsin(sin(46.0°) / 1.44))) / 1.44.
- Calculate the angle of refraction on the bottom surface: theta2 = arcsin(sin(180° - (90° - arcsin(sin(46.0°) / 1.44)) - (90° - arcsin(sin(46.0°) / 1.44))) / 1.44).
After following these steps, you should obtain the correct value for the angle at which the ray exits the prism, relative to the normal of that surface.