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

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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 ◦.

Ray diagram of a light ray entering a 26-64-90 degree glass prism normally, refracting at point P and again at the exit face; labels n = 1.69 for the glass and n = 1.333 for the water, angles 64°, 26°, θ1, θ3, θ4.


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[tex]\theta[/tex]1=n2sin[tex]\theta[/tex]2

The Attempt at a Solution



My work:

Handwritten working for a prism refraction problem: a ray enters a right-angled prism at 26 and 64 degree angles, with theta 3 = 180 - (116 - 21) = 43, n1 sin theta3 = n2 sin theta4, and sin theta4 = 0.5666, giving theta4 = 60.07, with n = 1.69 and n = 1.33 noted at left.


Handwritten work on a prism refraction problem: a ray leaves a boundary at 64 degrees with the normal and bends to the right, labelled glass = 1.69 and air = 1, with n1 sin theta1 = n2 sin theta2, 1.69 sin 64 = n2 sin 90, and a boxed result of 1.519 = n2.


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
  • Ray diagram of a light ray entering a 26-64-90 degree prism of glass with n = 1.69, immersed in water with n = 1.333, showing angles 64 degrees, theta 1, theta 3, theta 4 and 26 degrees, with point P on the upper face.
    DSC06483.jpg
    41.6 KB · Views: 552
  • Handwritten worked solution: a prism diagram with angles 26 and 64 degrees and a ray entering normally, then theta3 = 180 - (116 - 21) = 43 degrees, n1 sin theta3 = n2 sin theta4, 1.69 sin 43 = 1.33 sin theta4, giving theta4 = 60.07 degrees, with n = 1.69 and n = 1.33 noted at left.
    gjfgjfgjfgj.jpg
    21.9 KB · Views: 840
  • Handwritten work on a prism refraction problem: a ray meets a horizontal surface at 64° to the normal, labelled "air = 1" and "glass = 1.69", with n1 sin θ1 = n2 sin θ2 written below, then 1.69 sin 64 = n2 sin 90 giving boxed 1.519 = n2.
    gdsgdslo.jpg
    18.3 KB · Views: 670

Discussion

So it would be:

[tex]\theta[/tex]3=180-(26+26+90)=38[tex]\circ[/tex]

n3sin[tex]\theta[/tex]3=n4sin[tex]\theta[/tex]4
1.69sin38=1.33sin[tex]\theta[/tex]4
[tex]\frac{1.0405}{1.33}[/tex]=sin[tex]\theta[/tex]4
sin-10.7823=[tex]\theta[/tex]4
51=[tex]\theta[/tex]4

And #3 is right? TIA
 

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