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In summary, the conversation discusses the position of minimum deviation in a prism spectrometer and the angle shown in the attached image. It is mentioned that the angle of incidence equals the angle of reflection and that the minimum deviation can be found by rotating the table until the beam deviates the least. The derivation of the minimum angle of deviation is mentioned and it is noted that it is necessary to use an incident angle ## \alpha ## and an emerging angle ## \beta ## and set them equal to zero. It is also mentioned that the angle measured in the image is twice the angle of the prism, and that the deviation angle formula can be derived using Snell's law.

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Cutter Ketch

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VVS2000 said:In the image attached, How is it that the angle shown is 2A?

Keep in mind that the angle of incidence equals angle of reflection, and you should be able to do the trig yourself,

VVS2000 said:And what exactly is the position of min deviation? Why does it have to be in the given position as shown?

It doesn’t. They didn’t want to give it away. You are supposed to rotate the table until the beam deviates the least. You will find it does not occur in the orientation shown.

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VVS2000

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Well I was instructed to rotate the table until the image moves in the opposite direction of rotation. Why should this be the point where light deviates the least?Cutter Ketch said:Keep in mind that the angle of incidence equals angle of reflection, and you should be able to do the trig yourself,

It doesn’t. They didn’t want to give it away. You are supposed to rotate the table until the beam deviates the least. You will find it does not occur in the orientation shown.

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VVS2000

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No I was'nt talking about the derivation. When you're doing it practically, why does the angle which you measure in the picture I attached is 2A? Twice the angle of prism?Charles Link said:

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I was trying to answer your second question. I believe the first one is simply trigonometry.VVS2000 said:No I was'nt talking about the derivation. When you're doing it practically, why does the angle which you measure in the picture I attached is 2A? Twice the angle of prism?

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VVS2000

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Oh sry. Yeah I think I got the first one. I thought you were offering a different perspectiveCharles Link said:I was trying to answer your second question. I believe the first one is simply trigonometry.

Ok, so practically is'nt there anyway to convince that position is min deviation position?

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Deviation angle ## \Psi=\Psi(\alpha, n, A) ##, if I remember correctly. If you take ## \frac{\partial{\Psi}}{\partial{\alpha}}=0 ##, to find ## \Psi_{min} ##, I think a little computing gave the result that ## \alpha=\beta ## at the minimum. I could look for the lab manual tomorrow, but I don't think the derivation is a very difficult one. We should be able to derive it ourselves with a little effort, using Snell's law, etc. ## \\ ## Edit: With a little effort, I succeeded in doing the derivation of the formula. I would need to write it up neatly, (drawing a diagram, etc.), and take a photo of the derivation, etc., if you are interested. It is somewhat difficult to derive, but not tremendously so.

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VVS2000

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Ok I will give this a try.Charles Link said:Deviation angle ## \Psi=\Psi(\alpha, n, A) ##, if I remember correctly. If you take ## \frac{\partial{\Psi}}{\partial{\alpha}}=0 ##, to find ## \Psi_{min} ##, I think a little computing gave the result that ## \alpha=\beta ## at the minimum. I could look for the lab manual tomorrow, but I don't think the derivation is a very difficult one. We should be able to derive it ourselves with a little effort, using Snell's law, etc.

Thanks!

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See my edit to post 9.VVS2000 said:Ok I will give this a try.

Thanks!

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VVS2000

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Sure! If that's ok with you...Charles Link said:

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sophiecentaur

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This sort of question is much better answered by looking at an elementary optics textbook (or searching Google Images - which gives you many pictorial hits). Alternatively, just draw a diagram of a light ray, reflected in a mirror (a simpler model to start with) and see where it goes when you rotate the mirror a bit. If you don't like Maths too much, this should help. OR play with an actual mirror and light from a torch / LED pointer.VVS2000 said:No I was'nt talking about the derivation. When you're doing it practically, why does the angle which you measure in the picture I attached is 2A? Twice the angle of prism?

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sophiecentaur

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You can't say "why" something like that happens without a 'derivation'. The word may look like intimidating maths and something to steer clear of but the only alternative is to do many practical measurements with different angles and then you may come to the same conclusion that a few lines of maths will give you. (Only you can never be sure that the practical results apply generally and that's why we do sums in Physics)VVS2000 said:No I was'nt talking about the derivation. When you're doing it practically, why does the angle which you measure in the picture I attached is 2A? Twice the angle of prism?

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The following will apply to the figure of the above post.

Deviation angle ## \Psi=\alpha+\beta-\gamma ##.

also ## (90-\alpha')+(90-\beta')+\gamma=180 ##, so that ## \gamma=\alpha'+\beta' ##.

With Snell's law: ## \alpha'=\sin^{-1}(\sin(\alpha) /n) ##, so that ## \beta'=\gamma-\sin^{-1}(\sin(\alpha )/n) =\sin^{-1}(\sin(\beta)/n) ##.

There are just a few more steps.

We have ## \beta=\beta(\alpha) ##.

##\frac{d \Psi}{d \alpha}=0 ## gives ## 1+\frac{d \beta}{d \alpha}=0 ## at the minimum.

For ## \beta' ##, we have something of the form ## F(\beta)=\gamma-F(\alpha) ##.

This gives ##F'(\alpha) \, d \alpha=-F'(\beta) \, d \beta ##.

Since ## d\beta=-d \alpha ## at the minimum, we get ## F'(\alpha)=F'(\beta) ##,

and it should be fair to conclude that ## \alpha=\beta ## at the minimum.

Deviation angle ## \Psi=\alpha+\beta-\gamma ##.

also ## (90-\alpha')+(90-\beta')+\gamma=180 ##, so that ## \gamma=\alpha'+\beta' ##.

With Snell's law: ## \alpha'=\sin^{-1}(\sin(\alpha) /n) ##, so that ## \beta'=\gamma-\sin^{-1}(\sin(\alpha )/n) =\sin^{-1}(\sin(\beta)/n) ##.

There are just a few more steps.

We have ## \beta=\beta(\alpha) ##.

##\frac{d \Psi}{d \alpha}=0 ## gives ## 1+\frac{d \beta}{d \alpha}=0 ## at the minimum.

For ## \beta' ##, we have something of the form ## F(\beta)=\gamma-F(\alpha) ##.

This gives ##F'(\alpha) \, d \alpha=-F'(\beta) \, d \beta ##.

Since ## d\beta=-d \alpha ## at the minimum, we get ## F'(\alpha)=F'(\beta) ##,

and it should be fair to conclude that ## \alpha=\beta ## at the minimum.

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I'm going to complete the derivation using one more post:

Since ## \alpha=\beta ## at the minimum, we have

## \Psi_{min}=2 \alpha-\gamma ##,

and also ## 2 \sin^{-1}((\sin{\alpha})/n)=\gamma ##,

so that ## (\sin{\alpha})/n=\sin(\gamma /2) ##,

so that ##n=\frac{\sin{\alpha}}{\sin(\gamma /2)} ##.

Finally ## n=\frac{\sin((\Psi_{min}+\gamma)/2)}{\sin(\gamma /2)} ##.

Since ## \alpha=\beta ## at the minimum, we have

## \Psi_{min}=2 \alpha-\gamma ##,

and also ## 2 \sin^{-1}((\sin{\alpha})/n)=\gamma ##,

so that ## (\sin{\alpha})/n=\sin(\gamma /2) ##,

so that ##n=\frac{\sin{\alpha}}{\sin(\gamma /2)} ##.

Finally ## n=\frac{\sin((\Psi_{min}+\gamma)/2)}{\sin(\gamma /2)} ##.

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VVS2000

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Oh... ! Now it makes full sense! Thanks man! I Hope to return the favour sometime laterCharles Link said:

Since ## \alpha=\beta ## at the minimum, we have

## \Psi_{min}=2 \alpha-\gamma ##,

and also ## 2 \sin^{-1}((\sin{\alpha})/n)=\gamma ##,

so that ## (\sin{\alpha})/n=\sin(\gamma /2) ##,

so that ##n=\frac{\sin{\alpha}}{\sin(\gamma /2)} ##.

Finally ## n=\frac{\sin((\Psi_{min}+\gamma)/2)}{\sin(\gamma /2)} ##.

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It may interest you that in the use of a spectrometer, ##n=n(\lambda) ##, so that the ## \beta ## will be different between ## \lambda_1 ## and ## \lambda_2 ##, for a particular ## \alpha ##. If the spectrometer is set at the minimum ## \Psi_{min} ##, with the corresponding ## \alpha ## where ## \alpha=\beta ## for ##\lambda_1 ##, then ##n(\lambda_1) ## can be calculated from ## \Psi_{min}(\lambda_1) ##.

I was able to show ,[back in college many years ago=I was the T.A. (teaching assistant) for a course that did a laboratory experiment using the prism spectrometer along with the formula for the index n ], with a somewhat detailed calculation, that if you use the ## \Psi(\lambda_1) ## above to compute ##n( \lambda_1) ##, you can also, to a very good approximation, find ## n(\lambda_2) ##, using ## \Psi_{min}(\lambda_2) \approx \Psi_{min}(\lambda_1)+\Delta \beta ##, where ## \Delta \beta=\beta(\lambda_2)-\beta(\lambda_1) ##, with ## \alpha ## fixed at ##\alpha=\beta(\lambda_1) ## as above. It turns out that ## \Psi(\lambda_2) ## is also very near ## \Psi_{min}(\lambda_2) ##, so that ## \Delta \Psi_{min} \approx \Delta \beta ##, for ##\lambda_1 ## and ##\lambda_2 ##. It's a lot of detail, but you might find it of interest. ## \\ ## In using a prism spectrometer, it is of interest to be able to compute (and measure) ## \Delta \beta ## for two different wavelengths, from the graph of ## n(\lambda) ## vs. ## \lambda ##. This is possible, with the approximation ## \Delta \Psi_{min} \approx \Delta \beta ##, along with the formula they give you for ## n ## as a function of ## \Psi_{min} ##. ## \\ ## Note: To reach ## \Psi_{min}(\lambda_2) ##, it is necessary to change ## \alpha ## to a new location, so that once again ## \alpha=\beta ##, but this time for ## \lambda_2 ##. What essentially happens is the prism is rotated by an amount ## \Delta \theta=\Delta \beta /2 ##. In the process, ## \alpha ## increases, and ## \beta ## decreases by that amount, with ## \Psi ## remaining virtually unchanged, since near the minimum ## \frac{d \Psi}{d \alpha} \approx 0 ##. This may be a lot of detail, but you might find it of interest. The spectrum tends to function as a unit when the prism is rotated, with wavelengths that are near each other having ## \Delta \Psi_{min}=\Delta \beta ##, where the ## \beta ## is for some fixed ## \alpha ##, e.g. the ## \alpha ## that makes for ## \Psi_{min}(\lambda_1) ##. The ## \Psi(\lambda_2) ## was already very nearly at ## \Psi_{min}(\lambda_2) ##, even before this slight ## \Delta \theta ## rotation.

I was able to show ,[back in college many years ago=I was the T.A. (teaching assistant) for a course that did a laboratory experiment using the prism spectrometer along with the formula for the index n ], with a somewhat detailed calculation, that if you use the ## \Psi(\lambda_1) ## above to compute ##n( \lambda_1) ##, you can also, to a very good approximation, find ## n(\lambda_2) ##, using ## \Psi_{min}(\lambda_2) \approx \Psi_{min}(\lambda_1)+\Delta \beta ##, where ## \Delta \beta=\beta(\lambda_2)-\beta(\lambda_1) ##, with ## \alpha ## fixed at ##\alpha=\beta(\lambda_1) ## as above. It turns out that ## \Psi(\lambda_2) ## is also very near ## \Psi_{min}(\lambda_2) ##, so that ## \Delta \Psi_{min} \approx \Delta \beta ##, for ##\lambda_1 ## and ##\lambda_2 ##. It's a lot of detail, but you might find it of interest. ## \\ ## In using a prism spectrometer, it is of interest to be able to compute (and measure) ## \Delta \beta ## for two different wavelengths, from the graph of ## n(\lambda) ## vs. ## \lambda ##. This is possible, with the approximation ## \Delta \Psi_{min} \approx \Delta \beta ##, along with the formula they give you for ## n ## as a function of ## \Psi_{min} ##. ## \\ ## Note: To reach ## \Psi_{min}(\lambda_2) ##, it is necessary to change ## \alpha ## to a new location, so that once again ## \alpha=\beta ##, but this time for ## \lambda_2 ##. What essentially happens is the prism is rotated by an amount ## \Delta \theta=\Delta \beta /2 ##. In the process, ## \alpha ## increases, and ## \beta ## decreases by that amount, with ## \Psi ## remaining virtually unchanged, since near the minimum ## \frac{d \Psi}{d \alpha} \approx 0 ##. This may be a lot of detail, but you might find it of interest. The spectrum tends to function as a unit when the prism is rotated, with wavelengths that are near each other having ## \Delta \Psi_{min}=\Delta \beta ##, where the ## \beta ## is for some fixed ## \alpha ##, e.g. the ## \alpha ## that makes for ## \Psi_{min}(\lambda_1) ##. The ## \Psi(\lambda_2) ## was already very nearly at ## \Psi_{min}(\lambda_2) ##, even before this slight ## \Delta \theta ## rotation.

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A spectrometer is a scientific instrument used to measure the intensity of light at different wavelengths. It separates the different wavelengths of light using a prism or diffraction grating, and then measures the intensity of each wavelength.

The refractive index of a prism can be measured using a spectrometer by measuring the angle of deviation of a light beam passing through the prism. This angle can then be used to calculate the refractive index using Snell's law.

The refractive index of a prism is important because it determines the amount of light that is refracted or bent as it passes through the prism. This is essential in many scientific and technological applications, such as in optics and spectroscopy.

Several factors can affect the accuracy of measuring the refractive index of a prism using a spectrometer, including the quality of the prism, the precision of the spectrometer, and external factors such as temperature and air pressure.

Yes, the refractive index of a prism can vary for different wavelengths of light. This phenomenon is known as dispersion and is the reason why white light is separated into its component colors when passing through a prism.

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