What Angle Does Light Emerge from a Prism?

  • Thread starter Thread starter OhBoy
  • Start date Start date
  • Tags Tags
    Light Prism
AI Thread Summary
A light ray enters a prism with an index of refraction of 1.5 and is initially horizontal. The ray refracts at an angle of 19.47 degrees with the normal inside the prism and then strikes the second face. The calculated angle of emergence was initially found to be 77 degrees, but the correct answer is 47 degrees when considering the angle relative to the horizontal. The discussion emphasizes the importance of visualizing the ray's path and understanding the relationship between the angles and the prism's geometry. Ultimately, the correct approach involves applying Snell's Law and accurately determining the angles with respect to the horizontal.
OhBoy
Messages
29
Reaction score
0

Homework Statement


A light ray traveling in the horizontal direction is incident onto a prism as shown in the figure. At what angle relative to horizontal does the light ray emerge from the second face of the shown prism if the prism has an index of refraction of 1.5 and is surrounded by air? The cross section of the prism is in the shape of an equilateral triangle.

Homework Equations


Snells Law.

The Attempt at a Solution

Okay, if the light is heading into the left side of the triangle, then the angle it makes with the normal is 30 degrees since the angle the ray to the left side of the triangle would be 60 degrees. That means the refracted ray will make 19.47 degrees with the normal inside the prism. Now, I believe I should do 90 - 19.47 to get the angle that will make a triangle to connect me to the right side of the prism. Then I can do 180-(90-19.47)-60 = 30+19.47 which gets me 49.47 degrees. Now I need the angle normal to the right side so I do 90-49.47 and I get 40.53 degrees. Snells law tells me that the refracted ray on the right side should be arcsin(1.5*sin(40.53)), which is about 77 degrees. This answer is off by 30 degrees for some reason. The answer is 47 degrees but I get 77 degrees.
 
Physics news on Phys.org
Draw a rough sketch and see whether the refracted ray is meeting the opposite face of the prism or which face of the prism! Check the angle of 49.47 you have got is between refracted ray and which face of prism. Just do not play with numbers visualize what is happening.
 
Let'sthink said:
Draw a rough sketch and see whether the refracted ray is meeting the opposite face of the prism or which face of the prism! Check the angle of 49.47 you have got is between refracted ray and which face of prism. Just do not play with numbers visualize what is happening.

There's a ray of light going right on the triangle's left face horizontally. It is then refracted downwards and hits the right face. It is then further refracted downwards. The 49.47 degrees should be the angle between the ray of light and the right side of the triangle.
 
OhBoy said:
Snells law tells me that the refracted ray on the right side should be arcsin(1.5*sin(40.53)), which is about 77 degrees. This answer is off by 30 degrees for some reason. The answer is 47 degrees but I get 77 degrees.
That would be with respect to the local normal to the prism. What's the normal's angle with respect to the horizontal? (You were asked for the ray's angle with respect to the horizontal).
 
  • Like
Likes OhBoy
gneill said:
That would be with respect to the local normal to the prism. What's the normal's angle with respect to the horizontal? (You were asked for the ray's angle with respect to the horizontal).

Aha! Thank you! The being off by 30 degrees should of really gave it away. Thanks again for your help :)
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top