Uncertainties and the angle of minimum deviation

[Jane Smith]

Homework Statement

For a physics experiment I need to find the uncertainties and I am using the angle of minimum derivation formula:

The value of A=60° and one of the values of D is 29.7° which has a uncertainty of ±2 (I know it's a very high value)

Homework Equations

How do I calculate the uncertainty for this?

The Attempt at a Solution

I found a video which included some formulas but I did not understand (attached photos)

 

[berkeman]

You mean Angle of Minimum Deviation, right? https://en.wikipedia.org/wiki/Minimum_deviation

EDIT -- I went ahead and fixed your thread title for you.

 

[Charles Link]

When you write $n_{\lambda}=f(D_{\lambda} )$, then $\sigma_{n_{\lambda}}=(\frac{\partial{n_{\lambda}}}{\partial{D_{\lambda}}}) \sigma_{D_{\lambda}}$. $\\$ Do you know how to take the derivative? (In this case, you really only have one variable $D_{\lambda}$, so the partial derivative signs aren't necessary). $\\$ In the page that you show in the OP on the left, I think they are ignoring any uncertainties in the apex angle.

 

[Jane Smith]

When you write $n_{\lambda}=f(D_{\lambda} )$, then $\sigma_{n_{\lambda}}=(\frac{\partial{n_{\lambda}}}{\partial{D_{\lambda}}}) \sigma_{D_{\lambda}}$. $\\$ Do you know how to take the derivative? (In this case, you really only have one variable $D_{\lambda}$, so the partial derivative signs aren't necessary).

No

 

[Charles Link]

See the additional line I added above (Post 3). The derivative involves the chain rule. The derivative of the $\sin(\theta)$ is $\cos(\theta)$. On this problem, it is almost necessary to have had at least one calculus course to be able to see what they are doing.

 

[Charles Link]

If the value of $A=60^{\circ}$, I think that should be $30^{\circ}$ rather than $45^{\circ}$ inside the cosine function.$\\$ (I think they may have used $A=90^{\circ}$, and that would explain the factot of $\sqrt{2}$ that they get in the denominator, i.e. $\sin(\frac{90^{\circ}}{2})=\frac{\sqrt{2}}{2}$ ).

 

[Jane Smith]

See the additional line I added above (Post 3). The derivative involves the chain rule. The derivative of the $\sin(\theta)$ is $\cos(\theta)$. On this problem, it is almost necessary to have had at least one calculus course to be able to see what they are doing.

I did take calculus last year but its been so long I don't feel confident on my abilities. But once I did derive it how would I proceed from there. Would it be possible to calculate this graphically?

 

[Charles Link]

I can show you the calculus. Let $D_{\lambda}=x$. We will treat $A$ as a constant. The formula you have for $n_{\lambda}$ is then $n_{\lambda}=\frac{\sin(\frac{x}{2}+\frac{A}{2} )}{\sin(\frac{A}{2})}$. $\\$ $(\frac{dn_{\lambda}}{dx})=(\frac{1}{\sin(\frac{A}{2})})(\frac{1}{2})( \cos(\frac{x}{2}+\frac{A}{2}))$. $\\$ [And $\sigma_r=(\frac{dn_{\lambda}}{dx})(\sigma_L)$].$\\$ Can you follow the chain rule that I used to get the $(\frac{1}{2} )$, (middle term), because it is $\sin(\frac{x}{2}+\frac{A}{2} )$, (with a 2 below the $x$) ? $\\$ I'll show you the chain rule here: If $f(x)=\sin(x)$, then $f'(x)=\cos(x)$, $\\$ but if $f(x)=\sin(\frac{x}{2}+c)$, then $f'(x)=(\frac{df(u)}{du})(\frac{du}{dx} )$, where in this case $u=\frac{x}{2}+c$. $\\$ Meanwhile the factor $\sin(\frac{A}{2})$ in the denominator is simply a constant factor: i.e. $\frac{d( Cf(x))}{dx}= C \frac{d f(x)}{dx}$. $\\$ The calculus is a little easier than it first looks. Don't be afraid to study it=I do think you should be able to follow the derivation. $\\$ Edit: See also post 6 above again=I added some detail.

 

[Jane Smith]

I can show you the calculus. Let $D_{\lambda}=x$. We will treat $A$ as a constant. The formula you have for $n_{\lambda}$ is then $n_{\lambda}=\frac{\sin(\frac{x}{2}+\frac{A}{2} )}{\sin(\frac{A}{2})}$. $\\$ $(\frac{dn_{\lambda}}{dx})=(\frac{1}{\sin(\frac{A}{2})})(\frac{1}{2})( \cos(\frac{x}{2}+\frac{A}{2}))$. $\\$ [And $\sigma_r=(\frac{dn_{\lambda}}{dx})(\sigma_L)$].$\\$ Can you follow the chain rule that I used to get the $(\frac{1}{2} )$, (middle term), because it is $\sin(\frac{x}{2}+\frac{A}{2} )$, (with a 2 below the $x$) ? $\\$ I'll show you the chain rule here: If $f(x)=\sin(x)$, then $f'(x)=\cos(x)$, $\\$ but if $f(x)=\sin(\frac{x}{2}+c)$, then $f'(x)=(\frac{df(u)}{du})(\frac{du}{dx} )$, where in this case $u=\frac{x}{2}+c$. $\\$ Meanwhile the factor $\sin(\frac{A}{2})$ in the denominator is simply a constant factor: i.e. $\frac{d( Cf(x))}{dx}= C \frac{d f(x)}{dx}$. $\\$ The calculus is a little easier than it first looks. Don't be afraid to study it=I do think you should be able to follow the derivation. $\\$ See also post 6 above again=I added some detail.

Is this correct? How do I find the uncertainty using this?

 

[Charles Link]

The uncertainty is $\sigma_r$. They might do well to call it $\Delta n$, but they are using statistical type of language, where $\sigma$ is the letter they use for standard deviation. And yes, what you have is correct. The uncertainty $\Delta n=\sigma_r=(\frac{dn}{dx})\sigma_L=(\frac{dn}{dx}) \Delta_{\theta_L}$. $\\$ One additional note: This one is a little tricky, because when you put in $\Delta \theta_L= \sigma_L$ i this formula, you need to convert it to radians, i.e. express $\Delta \theta_L$ in radians. The result is $\Delta \theta_L=2 (\frac{\pi}{180} )$.Everywhere else you can use degrees, and it doesn't matter, but here you need to have $\Delta \theta_L$ in radians. $\\$ The reason for this is the derivative of the $\sin(x)$ is equal to the $\cos(x)$ only when $x$ is measured in radians. Otherwise there is an additional factor. (This is kind of an advanced topic. I'll try to explain:) To start this explanation, they are basically writing $dn=\Delta n=(\frac{dn}{d \theta_L}) \, d \theta_L=(\frac{dn}{d \theta_L}) \, \Delta \theta_L$. The derivative gives you the cosine only if it is measured in radians. The derivative derivation assumes $\frac{\sin(\Delta x)}{\Delta x }=1$ for small $\Delta x$, but that is only the case of $\Delta x$ is measured in radians. If $\Delta x$ is measured in degrees, then in taking the derivative, the factor that appears of $\frac{\sin(\Delta x)}{\Delta x}$ for small $\Delta x$ will give a factor of $\frac{\pi}{180}$ instead of a factor of one.

 

[Jane Smith]

The uncertainty is $\sigma_r$. They might do well to call it $\Delta n$, but they are using statistical type of language, where $\sigma$ is the letter they use for standard deviation. And yes, what you have is correct. The uncertainty $\Delta n=\sigma_r=(\frac{dn}{dx})\sigma_L=(\frac{dn}{dx}) \Delta_{\theta_L}$. $\\$ One additional note: This one is a little tricky, because when you put in $\Delta \theta_L= \sigma_L$ i this formula, you need to convert it to radians, i.e. express $\Delta \theta_L$ in radians. The result is $\Delta \theta_L=2 (\frac{\pi}{180} )$.Everywhere else you can use degrees, and it doesn't matter, but here you need to have $\Delta \theta_L$ in radians. $\\$ The reason for this is the derivative of the $\sin(x)$ is equal to the $\cos(x)$ only when $x$ is measured in radians. Otherwise there is an additional factor. (This is kind of an advanced topic. I'll try to explain:) To start this explanation, they are basically writing $dn=\Delta n=(\frac{dn}{d \theta_L}) \, d \theta_L=(\frac{dn}{d \theta_L}) \, \Delta \theta_L$. The derivative gives you the cosine only if it is measured in radians. The derivative derivation assumes $\frac{\sin(\Delta x)}{\Delta x }=1$ for small $\Delta x$, but that is only the case of $\Delta x$ is measured in radians. If $\Delta x$ is measured in degrees, then in taking the derivative, the factor that appears of $\frac{\sin(\Delta x)}{\Delta x}$ for small $\Delta x$ will give a factor of $\frac{\pi}{180}$ instead of a factor of one.

Okay so, since 29.7° is equal to 0.5183628 radians. Do I simply just use that in the formula.

To which I get the answer 0.7, meaning there is an uncertainty of ±0.7?

 

[Charles Link]

The only place you need to convert to radians is in computing $\sigma_L=\Delta \theta_L=2(\frac{\pi}{180})$. Everywhere else, simply use degrees and set your calculator to "degrees". $\\$ The uncertainty in the refractive index $\Delta n=\sigma_n=(\frac{dn}{dD_{\lambda}})\sigma_L$. I think you are saying you get a factor of .7 for the derivative (which is also what I get), but you need to multiply by $\sigma_L \approx .035$ (radians) to get the uncertainty $\sigma_n=\Delta n$. $\\$ I don't know if they specify in the notes they gave you that $\sigma_L= \Delta \theta_L$ needs to be measured in radians in your formula for $\sigma_n \, (=\sigma_r)$, but this is very important. Without it, the results are incorrect= you get uncertainty in $n$ that is $\Delta n=\pm 1.4$. (Remember $\sigma_L=2$ if you incorrectly write it in degrees in this formula). $\\$ If you compute the $\Delta \theta_L$ in radians (the correct way to do it), you get $\Delta n=\pm .02$. $\\$ The index $n$ is normally about 1.5 for glass, and with an uncertainty of $\Delta n=\pm 1.4$, it would be a very useless experiment. Fortunately, the correct uncertainty $\Delta n =\pm .02$, and it is actually a very accurate experiment.

 

[Jane Smith]

The only place you need to convert to radians is in computing $\sigma_L=\Delta \theta_L=2(\frac{\pi}{180})$. Everywhere else, simply use degrees and set your calculator to "degrees". $\\$ The uncertainty in the refractive index $\Delta n=\sigma_n=(\frac{dn}{dD_{\lambda}})\sigma_L$. I think you are saying you get a factor of .7 for the derivative (which is also what I get), but you need to multiply by $\sigma_L \approx .035$ (radians) to get the uncertainty $\sigma_n=\Delta n$. $\\$ I don't know if they specify in the notes they gave you that $\sigma_L= \Delta \theta_L$ needs to be measured in radians in your formula for $\sigma_n \, (=\sigma_r)$, but this is very important. Without it, the results are incorrect= you get uncertainty in $n$ that is $\Delta n=\pm 1.4$. (Remember $\sigma_L=2$ if you incorrectly write it in degrees in this formula). $\\$ If you compute the $\Delta \theta_L$ in radians (the correct way to do it), you get $\Delta n=\pm .02$. $\\$ The index $n$ is normally about 1.5 for glass, and with an uncertainty of $\Delta n=\pm 1.4$, it would be a very useless experiment. Fortunately, the correct uncertainty $\Delta n =\pm .02$, and it is actually a very accurate experiment.

I am sorry, but I am confused to how you get to 0.035

The only place you need to convert to radians is in computing $\sigma_L=\Delta \theta_L=2(\frac{\pi}{180})$. Everywhere else, simply use degrees and set your calculator to "degrees". $\\$ The uncertainty in the refractive index $\Delta n=\sigma_n=(\frac{dn}{dD_{\lambda}})\sigma_L$. I think you are saying you get a factor of .7 for the derivative (which is also what I get), but you need to multiply by $\sigma_L \approx .035$ (radians) to get the uncertainty $\sigma_n=\Delta n$. $\\$ I don't know if they specify in the notes they gave you that $\sigma_L= \Delta \theta_L$ needs to be measured in radians in your formula for $\sigma_n \, (=\sigma_r)$, but this is very important. Without it, the results are incorrect= you get uncertainty in $n$ that is $\Delta n=\pm 1.4$. (Remember $\sigma_L=2$ if you incorrectly write it in degrees in this formula). $\\$ If you compute the $\Delta \theta_L$ in radians (the correct way to do it), you get $\Delta n=\pm .02$. $\\$ The index $n$ is normally about 1.5 for glass, and with an uncertainty of $\Delta n=\pm 1.4$, it would be a very useless experiment. Fortunately, the correct uncertainty $\Delta n =\pm .02$, and it is actually a very accurate experiment.

I think I finally understand, thank you so much :)

 

[Charles Link]

Just one additional comment: It would be a very easy mistake for the professor to make if he forgot to include the detail that the uncertainty in the measured angle needs to be in radians in using it to compute $\Delta n$. The "radians" is actually dimensionless just like the index of refraction is, so that both sides of the equation $\Delta n=(\frac{d n}{d D_{\lambda}}) \Delta \theta_L$ are free of units. $\\$ In any case, it is very important that the number $\Delta \theta_L$ is expressed in radians in the calculation of $\Delta n$, because of some finer details about the derivative of $\sin(x)$ that are discussed in post 10 above. $\\$ ===========================================================================$\\$ Additional item: If you have have ever seen the Taylor series: $\\$ $\sin(x)=x-\frac{x^3}{3!} + \frac{x^5}{5!} +...$, $\\$ you might ask, why does $x$ need to be measured in radians for this formula to work? $\\$ It comes from $f(x)=f(a)+f'(a)(x-a)+f''(a) \frac{(x-a)^2}{2!} +...$.$\\$ The reason the radians are required is because the derivatives are only correct when $x$ is measured in radians. If $x$ is measured in degrees, the derivatives pick up a factor of $\frac{\pi}{180}$ as mentioned in post 10 above.

 

[Ray Vickson]

Homework Statement

For a physics experiment I need to find the uncertainties and I am using the angle of minimum derivation formula:

The value of A=60° and one of the values of D is 29.7° which has a uncertainty of ±2 (I know it's a very high value)

Homework Equations

How do I calculate the uncertainty for this?

The Attempt at a Solution

I found a video which included some formulas but I did not understand (attached photos)View attachment 237048 View attachment 237050

The very simplest way is to take the three values $D = 29.7$ (central value), $D_{low} = 29.7 - 2 = 27.7$ (lowest value) and $D_{hi} = 29.7 + 2 = 31.7$ (highest value), then do the calculation for all three values of $D$.

 

[Charles Link]

The very simplest way is to take the three values $D = 29.7$ (central value), $D_{low} = 29.7 - 2 = 27.7$ (lowest value) and $D_{hi} = 29.7 + 2 = 31.7$ (highest value), then do the calculation for all three values of $D$.

i.e. Calculate $n$ for the 3 values of $D$. Yes, this is a good way, and should also get the result that $\Delta n \approx \pm .02$.