Deviation of wavelengths through barium crown pism

• Liquidxlax
In summary, the refractive index of an equilateral prism of dense barium crown glass varies with wavelength as given in the table. Using Cauchy's dispersion formula, determine the minimum angle of deviation for sodium light with wavelength = 589.3 nm.
Liquidxlax

Homework Statement

The refractive index of an equilateral prism of dense barium crown glass varies with wavelength as given in the table

wavelength (nm)...n

656.3...1.635
486.1...1.646

making use of Cauchy's dispersion formula determine the minimum angle of deviation for sodium light with wavelength = 589.3 nm

Homework Equations

A/2 = theta prime

nsin(theta prime) = sin((A + deviation)/2)

(weird D)/(deviation) = (nf - nc)/(nd-1)

The Attempt at a Solution

2sin^-1((1.635)sin(30)) - 60 = deviation = 49.7 degrees

2sin^-1((1.646)sin(30)) - 60 = deviation = 50.8 degrees

weird D = 50.8 - 49.7 = 11.1 degrees

My problem is that I'm not sure how to find nd to finish the question.

i know i can use

n of wavelength = A + B/(wavelength)^2 + C/(wavelength)^4

but that doesn't leave me with an n between 1.635 and 1.646 and that is my problem

You can omit the third term C/(lambda)^4. Find A and B for the two refractive index - wavelength pairs.

ehild

ehild said:
You can omit the third term C/(lambda)^4. Find A and B for the two refractive index - wavelength pairs.

ehild

well I'm not totally sure on this but does this look right, or am i assuming the wrong thing

1.646 = 1.635 + b/((486.3nm)^2)

b = 5.9122x10^-15

then

n = 1.635 + (5.9122x10^-15)/((589.3x10^-9)^2) = 1.652 , but that is greater than 1.646 which is probably wrong then... I really don't know how to solve for A and B. I looked up their values for barium crown glass and it still doesn't work even with those

Liquidxlax said:
well I'm not totally sure on this but does this look right, or am i assuming the wrong thing

1.646 = 1.635 + b/((486.3nm)^2)

b = 5.9122x10^-15

No, A is not equal to 1.635, why should it?

1.635=A + B/656.3^2

and

1.646 = A + B/486.1^2.

Two equations, two unknowns. Solve.

ehild

ehild said:
No, A is not equal to 1.635, why should it?

1.635=A + B/656.3^2

and

1.646 = A + B/486.1^2.

Two equations, two unknowns. Solve.

ehild
:facepalm: why didn't i think of that... homework is 3x harder when you're sick

Okay i did that and i got a really high number, it seems unreasonable for an angle of deviation, but my textbook examples show ratios of 1/29 1/40... i got for the deviation

643.8 degrees

Show what did you get for A and B.

ehild

n1 = A + B/lambda1^2 represent A as a function of B A = n1 - B/lambda1^2
plug this in for A in the second one

n2 = A + b/lambda2^2 then n2 = (n1 - B/lambda1^2) + B/lambda2^2

n2-n1 = B(1/lambda2^2 - 1/lambda1^2) get B = (5.758x10^-15)

Then plug B into either equation solve for A to get A = (1.6216)

Then

n3 = 1.6216 + (5.758x10^-15)/(589.3x10^-9)^2 = 1.638

then plug the rest of the numbers in and solve for deviation in

n3 = nd n1 = nc n2 = nf

(weird D)/(deviation) = (nf-nc)/(nd-1)

deviation = ((weird D)(nd -1))/(nf-nc) = 643.8

I do not understand what you did. The question was:

"making use of Cauchy's dispersion formula determine the minimum angle of deviation for sodium light with wavelength = 589.3 nm". You calculated the minimum deviation for the other wavelengths with the equation

2sin^-1(n sin(30)) - 60 = deviation

why do not do the same for 589.3 nm? It is about 50 degrees.

ehild

i'll try that, it seems to make more sense, yet i did the lab on this today and they did it the way i initially was using. But that was a slightly different process.

1. What is barium crown prism and how does it affect wavelengths?

Barium crown prism is a type of prism made from barium crown glass, a high refractive index material. When light passes through a barium crown prism, it is refracted or bent, causing the different wavelengths of light to separate and form a spectrum.

2. How does the deviation of wavelengths through barium crown prism differ from other types of prisms?

The deviation of wavelengths through a barium crown prism is different from other types of prisms because it has a higher refractive index, meaning it can bend light at a greater angle. This results in a larger separation of wavelengths and a more distinct spectrum.

3. Can the deviation of wavelengths through barium crown prism be manipulated?

Yes, the deviation of wavelengths through barium crown prism can be manipulated by changing the angle of incidence, or the angle at which light enters the prism. By adjusting this angle, the amount of deviation can be increased or decreased.

4. What applications does the deviation of wavelengths through barium crown prism have?

The deviation of wavelengths through barium crown prism has various applications in fields such as spectroscopy, astronomy, and telecommunications. It is used to analyze the composition of materials, study the properties of light, and manipulate signals in fiber optic communication.

5. Are there any limitations to the deviation of wavelengths through barium crown prism?

One limitation of the deviation of wavelengths through barium crown prism is that it only works for a specific range of wavelengths. If the light source emits wavelengths outside of this range, they will not be separated by the prism. Additionally, the precision and accuracy of the deviation may be affected by factors such as the quality of the prism and external environmental conditions.

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