Diffraction Grating Homework: Zeroth & 1st Order Bright Spots

[aChordate]

Homework Statement

You shine a green laser pointer (l = 535 nm) on a diffraction grating that contains 1.0
x 104 lines/cm. When projected onto a screen that is 5.0 m away from the grating, how far
apart will the zeroth-order and first-order bright spots be?

Homework Equations

sinθ = m λ/d

The Attempt at a Solution

sinθ = 0 * 535nm/0.009615cm
θ =0

sinθ = 1 * 535nm/0.009615cm
θ = 0.3188

tanθ=y/L
y = Ltanθ=5mtan0.3188 = 0.0278 m

 

[Simon Bridge]

Why is it $d\sin\theta = m\lambda$ for a diffraction grating but $d\sin\theta = (m+\frac{1}{2})\lambda$ for a double slit in your other thread?
I hope you converted the units before doing the calculations ...
Are the dimensions of the setup sufficient to use an approximation to get the distances more directly?

 

[technician]

The equation you have used is correct and if you get the value of Sinθ then you can find θ straight off.
I have not checked your units but I would say that it is bad practice to mix units in an equation.
You have nm and cm together!...I would change all units to m.

 

[aChordate]

Why is it $d\sin\theta = m\lambda$ for a diffraction grating but $d\sin\theta = (m+\frac{1}{2})\lambda$ for a double slit in your other thread?
I hope you converted the units before doing the calculations ...
Are the dimensions of the setup sufficient to use an approximation to get the distances more directly?

I am not sure why one equation is for diffraction grating and the other is for the double slit experiment. I think you would be more qualified to answer that.

Are the dimensions of the setup sufficient to use an approximation to get the distances more directly?

I'm not sure what you mean. Are you asking if the equation is appropriate for this question?

 

[Simon Bridge]

I am not sure why one equation is for diffraction grating and the other is for the double slit experiment. I think you would be more qualified to answer that.

Be that as it may, my knowledge is not what is important here.

I have asked why you chose which equation because you did the question and presumably you had to make a decision. I am asking about your decision making process - which I am not more qualified to answer - which will tell me how you think about physics problems - which will help me tailor an answer to how you think - so you will be more likely to understand it - so I know I'm not wasting your time. See?

Note: I don't need you to be sure ;)

Are the dimensions of the setup sufficient to use an approximation to get the distances more directly?
I'm not sure what you mean. Are you asking if the equation is appropriate for this question?

There is another equation from the one you have used which gets you the answer quicker - but it is only valid in specific circumstances.
The equation you have used is also an approximation - is the approximation valid? Could have used a different equation?

You can get the answer using your approach here - but you need to say more about how you are thinking about the problem if anyone is to help you.
It also helps if you actually ask a question: all you did was lay out a bunch of raw working with no discussion or any indication of what the problem was or if there is a problem at all. I know it feels safer to just write the least possible - less risk of appearing silly. Relax. We've all been there, we understand.

I suspect you are taking the approach of memorizing equations and applying them without understanding the physics.

 

[technician]

To try to help...I would strongly suggest that you check your units.
Your numerical answer is incorrect because of your mixed units.
You need to sort this out before you worry too much about which equation to use.
The equation you are using is a standard textbook equation.

 

[Simon Bridge]

Your numerical answer is incorrect because of your mixed units.

if the units were not converted, then the sine would end up bigger than one, and the inverse sine would be complex. I'd expect a digital calculator would return an error in that case. OP did not get that result - so, presumably, something else happened.

Certainly the second step is where there is a mistake.
One can convert the cm measurement into nm to match them up.
i.e. what happens if you divide nanometers by centimeters?

 

[technician]

If units are converted to metres the problems of dividing nanometres by centimetres does not arise.
Mixing units is bad practice in my opinion.

 

[Simon Bridge]

If you convert the units correctly, do you get the same number as above?

Converting everything to SI units makes sure of consistent results - but, here, you only need the same units in numerator and denominator. Recall that the word "nanometer" is just another way to write "10^-9 meters" and "centimeter" another way to write "10^-2 meters"... though it does help to make the exponent explicit.

But I should have been more clear - the question was intended for OP.

 

[ehild]

Why is it $d\sin\theta = m\lambda$ for a diffraction grating but $d\sin\theta = (m+\frac{1}{2})\lambda$ for a double slit in your other thread?

In case of the diffraction grating the distance of maxima was the question. The other thread (Young experiment) was about the distance between the dark fringes.

ehild

 

[Simon Bridge]

aChordata has a history of using PF to check homework ... that's OK, but it shows a need to learn how to do the checking without us. To this end, I have tried to ask guiding questions and would encourage others to do the same.

Part of that process is to make explicit the process of doing the working and asking questions about it.
I had hoped that aChordata would answer the questions before I continued.
Of course, this is not always what people want from PF :)