Exit Angle & N2: Solving 26°-64°-90° Prism Problem

  • Thread starter Thread starter DrMcDreamy
  • Start date Start date
  • Tags Tags
    Angle
AI Thread Summary
The discussion revolves around solving a two-part physics problem involving a light ray passing through a 26°–64°–90° prism made of dense flint glass immersed in water. The first part requires calculating the exit angle θ4 of the light ray, with the user expressing confusion over their initial calculations, which were deemed incorrect. They provided their method, which included determining θ3 and applying Snell's law, but received feedback suggesting their final angle calculation may need reevaluation. The second part of the problem involves finding the refractive index n2 at which total internal reflection ceases at a specific point in the setup. Overall, the user seeks clarification and guidance on their approach to both parts of the problem.
DrMcDreamy
Messages
68
Reaction score
0

Homework Statement



This is a two part problem:

2) As shown in the figure, a light ray is incident normally on one face of a 26◦–64◦–90◦ block of dense flint glass (a prism) that immersed in water. Find the exit angle θ4 of the light ray (Assume the index of grass is 1.69, and that of water is 1.333.) Answer in units of ◦.

attachment.php?attachmentid=32011&stc=1&d=1297117641.jpg


3) A substance is dissolved in the water to increase the index of refraction. At what value of n2 does total internal reflection cease at point P ?

Homework Equations



n1sin\theta1=n2sin\theta2

The Attempt at a Solution



My work:

attachment.php?attachmentid=32012&stc=1&d=1297117724.jpg


attachment.php?attachmentid=32013&stc=1&d=1297117921.jpg


I put in the answer to number 2 and it said its wrong. What I am doing wrong? Can you guys guide me through the problem? TIA!
 

Attachments

  • DSC06483.jpg
    DSC06483.jpg
    41.6 KB · Views: 513
  • gjfgjfgjfgj.jpg
    gjfgjfgjfgj.jpg
    21.9 KB · Views: 797
  • gdsgdslo.jpg
    gdsgdslo.jpg
    18.3 KB · Views: 629
Physics news on Phys.org
Anybody?!
 
In problem 2, theta3 = 180 - (26 + 26 + 90)
 
So it would be:

\theta3=180-(26+26+90)=38\circ

n3sin\theta3=n4sin\theta4
1.69sin38=1.33sin\theta4
\frac{1.0405}{1.33}=sin\theta4
sin-10.7823=\theta4
51=\theta4

And #3 is right? TIA
 
*bump*
 
It looks OK.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top