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

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In summary, the conversation is about a two-part problem involving a light ray incident on a triangular prism of dense flint glass immersed in water. The first part asks for the exit angle of the light ray, while the second part involves a substance being dissolved in the water to change the index of refraction and determining the value at which total internal reflection ceases. The attempted solution involves using the equation n1sinθ1=n2sinθ2 to find the exit angle in problem 2 and determining that the answer for problem 3 is correct. The person asking for help is unsure of what they are doing wrong and is seeking guidance.
  • #1
DrMcDreamy
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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 ◦.

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

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!
 

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  • #2
Anybody?!
 
  • #3
In problem 2, theta3 = 180 - (26 + 26 + 90)
 
  • #4
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
 
  • #5
*bump*
 
  • #6
It looks OK.
 

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

1. What is the "Exit Angle & N2" problem?

The "Exit Angle & N2" problem involves finding the exit angle and the refractive index (N2) of a prism when the input angle is known. It is commonly used in optics and physics experiments.

2. What is the importance of solving the 26°-64°-90° prism problem?

The 26°-64°-90° prism problem is important because it is a fundamental concept in optics and is used in various applications, such as designing optical instruments, calculating light refraction in different media, and understanding the behavior of light in different materials.

3. How do you solve the 26°-64°-90° prism problem?

To solve the 26°-64°-90° prism problem, you need to use the Snell's law which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the refractive indices of the two media. By rearranging this equation, you can find the exit angle and N2 of the prism.

4. What are the factors that affect the exit angle and N2 of a prism?

The exit angle and N2 of a prism are affected by the input angle, the refractive index of the first medium, and the refractive index of the prism material. These factors can also change depending on the wavelength of light being used.

5. How can the 26°-64°-90° prism problem be applied in real-life situations?

The 26°-64°-90° prism problem can be applied in various real-life situations, such as designing optical devices like cameras and microscopes, calculating the refraction of light through different materials, and understanding the behavior of light in different media like water and glass. It is also used in the field of optics and photonics for research and development purposes.

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