Waves -- Find the lowest possible values for m_r and m_v

[abm77]

Homework Statement

Two slits are separated by a distance of 4.40x10^-6m and illuminated with two monochromatic light sources with wave lengths of 600nm (red) and 400nm (violet). The m_r bright fringe of the red light coincides with the m_v bright fringe of the violet light. What are the lowest possible values for m_r and m_v?

d= 4.40x10^-6 m
λ_r = 600nm
λ_v = 400nm
m_r = ?
m_v = ?

Homework Equations

Δx/ L = λ/d
x_m = mLλ / d
d sinΘ_m = mλ

The Attempt at a Solution

I honestly don't know where to start as it seems I'm missing needed information for use any of the equations, but it is the last question of my assignment. Sorry if formatting is wrong, first time posting.

 

[Simon Bridge]

Welcome to PF;
When in doubt - draw a picture of what is happening.
Have you tried working out where the fringes go for each color of light ... sketch a diagram for the first 8 or so. Use the information you have to choose the equation you need.

 

[abm77]

Welcome to PF;
When in doubt - draw a picture of what is happening.
Have you tried working out where the fringes go for each color of light ... sketch a diagram for the first 8 or so. Use the information you have to choose the equation you need.

See that is what I would normally try and do to solve a question I'm struggling with, but let's just say my teacher isn't the best and is rushing through the unit and hasn't even showed us what everything is suppose to look like and has just thrown numbers and equations at us. Very frustrating way to learn.

 

[haruspex]

See that is what I would normally try and do to solve a question I'm struggling with, but let's just say my teacher isn't the best and is rushing through the unit and hasn't even showed us what everything is suppose to look like and has just thrown numbers and equations at us. Very frustrating way to learn.

You quote three equations. At least one of them looks very useful. You can write it out separately for the red and violet cases, and the only extra unknown is the same in both equations.

 

[abm77]

You quote three equations. At least one of them looks very useful. You can write it out separately for the red and violet cases, and the only extra unknown is the same in both equations.

I see that for the third equation that the extra unknown would be sinΘ_m, but I don't see how I can find that unknown value without knowing m, which is what I need to find anyways.
Edit: Maybe I'm missing something very obvious because I'm tired...

 

[haruspex]

I see that for the third equation that the extra unknown would be sinΘ_m, but I don't see how I can find that unknown value without knowing m, which is what I need to find anyways.
Edit: Maybe I'm missing something very obvious because I'm tired...

The unknown it introduces is just an angle. The m it corresponds to will be different for r and v, but it's the same angle. Write out the two versions of the equation, one for r and one for v, but use the same variable (just theta, say) for the angle.

 

[Simon Bridge]

The idea is to use what you know about maths and physics to deduce the correct picture.

You've seen what a fringe pattern looks like and you know how to get it.
haruspex has the standard approach - you write out the equations for mr and mv, and you know you need these values so that the fringes overlap. This will be when the two angles are the same.

If you are having a problem understanding how that would work, then go back a step ... write out the equation for $d\sin\theta_r = m_r\lambda_r$ right?
Solve it for $\theta_r$, and find values of $\theta_r$ for $m_r=1,2,3,4,\cdots$ ... repeat for $m_v$ and see which values are the same.

 

[abm77]

Thank you both, I get it now and completed the question. Don't know why it took me so long to understand, but I get it now.

 

[Simon Bridge]

There's a paralysis that hits some people when they are faced with having to start working on a question without knowing how to find the answer.
As you advance, these situations will crop up more and more. The secret is to play around - plug some numbers in etc.
Not every effort needs to be directed at the goal and you don't have to include everything you did in the answer you hand in.
It's less efficient right away but leads to solid intuitions that save time later when you need them to.