Helo with with a question about light please

  • #1

Homework Statement


A diffraction grating is labelled 300 lines mm^-1 (lines per mm) is put 120 cm from a screen. there is put white light on it.

The question is that i have to solve what the distance is between the m0 and the blue(450 nm) of m1.


Homework Equations


my formula : sin(ø)=(n*λ)/d



The Attempt at a Solution


I think that u have to find the angel between m0 and m1 blue but i just dont get it..

i have tried to do it this way but seems that it keeps turning out wrong.

sin(ø)=(n*λ)/d
sin(ø) is the angle i have to find
n is 1 since i have to find the distance from 0 to 1
λ is the wavelength 450 nm
and d must be 300 lines/mm --> D=3,33*10^-6

but from here i see everything i have done like wrong and then i dont understand it anymore

please help me, and by the way i am new here so sorry if i didnt put the questions the right place or such im still trying to learn it

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Doc Al
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You have the correct formula. So you should be able to find the angle and then the distance on the screen. What's the problem?
 
  • #3
You have the correct formula. So you should be able to find the angle and then the distance on the screen. What's the problem?

its because i am unsure about what the λ is in this one......if i have to use it the way i used it in the rest of my homework it should then be the the 450 nm of the blue, but i think that there maybe is a number you hhave to add on top of it ?

its because i have to have the distance between m0 and m1 and i think that i also have to + the 450 to that distance but seems that i am jsut thinking wrong or way to complicated ?

when using the formula i also get that the angle is 1,35135*10^-4 and then i don't know what to do from there
 
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  • #4
Doc Al
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its because i am unsure about what the λ is in this one......if i have to use it the way i used it in the rest of my homework it should then be the the 450 nm of the blue, but i think that there maybe is a number you hhave to add on top of it ?
The wavelength needed is the one given: 450 nm. When you do your calculation, be sure to express that in meters.

its because i have to have the distance between m0 and m1 and i think that i also have to + the 450 to that distance but seems that i am jsut thinking wrong or way to complicated ?
Once you get the angle, use a bit of right angle trig to get the distance on the screen.

when using the formula i also get that the angle is 1,35135*10^-4 and then i don't know what to do from there
Show how you made that calculation. (Is that sinθ or θ? Either way, it's way too small.)
 
  • #5
HallsofIvy
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You have
[tex]\frac{n\lambda}{d}= \frac{1(450 nm)}{3.33 \times 10^{-6}}[/tex]
450/3.33= 450(10/3)= 1500. The additional digits you have are "round off error" from writing (10/3) as 3.33. Since "300 lines per mm" is 1/300 mm per line= 10/3 x 10 ^{-5}m per line (not. "-6"). Also,be careful about the units. In the numerator, you have "nm" (10^{-9} m) and in the denominator "m". Since that is equal to [itex]sin(\theta)[/itex], after finding that value, take the "arcsin" of it.
 
  • #6
You have
[tex]\frac{n\lambda}{d}= \frac{1(450 nm)}{3.33 \times 10^{-6}}[/tex]
450/3.33= 450(10/3)= 1500. The additional digits you have are "round off error" from writing (10/3) as 3.33. Since "300 lines per mm" is 1/300 mm per line= 10/3 x 10 ^{-5}m per line (not. "-6"). Also,be careful about the units. In the numerator, you have "nm" (10^{-9} m) and in the denominator "m". Since that is equal to [itex]sin(\theta)[/itex], after finding that value, take the "arcsin" of it.


so since 1 nanometer = 1.0 × 10-9 meters i have to say 490*10^-9 =4,9*10^-7
but then i have the bottom that is still 3,33...*10^-6 because i said 1/300 can i just change it to 3,33....*10^-7 ?
 
  • #7
Doc Al
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so since 1 nanometer = 1.0 × 10-9 meters i have to say 490*10^-9 =4,9*10^-7
Sure, but 450 not 490.
but then i have the bottom that is still 3,33...*10^-6 because i said 1/300 can i just change it to 3,33....*10^-7 ?
Why would you think you can change it by a factor of ten??

d = (1 mm)/300 = (10^-3 m)/300 = 3.33 10^-6 m
 
  • #8
The wavelength needed is the one given: 450 nm. When you do your calculation, be sure to express that in meters.


Once you get the angle, use a bit of right angle trig to get the distance on the screen.


Show how you made that calculation. (Is that sinθ or θ? Either way, it's way too small.)

i made the calculation this way:

sinø=(1*(450*10^-9))m/1:300
 
  • #9
Sure, but 450 not 490.

Why would you think you can change it by a factor of ten??

d = (1 mm)/300 = (10^-3)/300 = 3.33 10^-6

because now i have the 450*10^-7 and the 3,33*10^-6 am i able to just devide them with each other when the one is -7 and the other is -6 ? if i am then should it look like this
and then the result is 13,51 and that would still be wrong ?

ps. thanks alot because u are using your time to help me it's really hard to understand how everything with the units work and very depressing since i know all the formulars yet the result only gets right with the right units
 
  • #10
Doc Al
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because now i have the 450*10^-7
That should be 450 nm = 450*10^-9 m.

(And yes, you can just divide them.)
 
  • #11
so the result on how to do it must then be 450*10^-7/(3,33*10^-6) = 13,51
13,51 must then be in meter right ?

after that i have to use some angle trig to solve the angle
and i think it is tan^-1 13,51 = 85,76

am i right so far ?

but since it is only m1 how come that the angle gets to big ? 85 seems to big i think
 
  • #12
Doc Al
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so the result on how to do it must then be 450*10^-7/(3,33*10^-6) = 13,51
13,51 must then be in meter right ?
Did you read the last post? 450*10^-7 is wrong; it should be 450*10^-9.
 
  • #13
Did you read the last post? 450*10^-7 is wrong; it should be 450*10^-9.

450*10^-9/(3,33*10^-6) = 0,1351 ? it doesn't give sense why you can divide the nanometer with ^-6 is it just a rule you have to remember ?

and shall i thereafter take that result and use tan^'1 on it or sin^-1 on it ? in an example i have in my book it says tan but i think it has to be sin^-1*(0,1351)= 7,76 must be the answer ?
 
  • #14
Doc Al
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450*10^-9/(3,33*10^-6) = 0,1351 ?
Right.
it doesn't give sense why you can divide the nanometer with ^-6 is it just a rule you have to remember ?
Not sure what you're talking about here. The prefix nano, as in nanometer (nm), means 10^-9.

and shall i thereafter take that result and use tan^'1 on it or sin^-1 on it ? in an example i have in my book it says tan but i think it has to be sin^-1*(0,1351)= 7,76 must be the answer ?
To find the angle, use sin^-1. But realize that for small enough angles, it doesn't matter which you use.

That's the angle, now you must use it to find the distance on the screen. That will involve tanθ.
 
  • #15
Right.

Not sure what you're talking about here. The prefix nano, as in nanometer (nm), means 10^-9.


To find the angle, use sin^-1. But realize that for small enough angles, it doesn't matter which you use.

That's the angle, now you must use it to find the distance on the screen. That will involve tanθ.

so until now i am right with the 7,76 ?
 
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  • #17
the distance... isn't it just d = (1*450*10^-9)/sin(7,76 )

i think it is this way and then the result will be 3,3327 *10^-6 meter

that distance seems very small i guess
 
  • #18
Doc Al
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the distance... isn't it just d = (1*450*10^-9)/sin(7,76 )
No. Once you have the angle, set up a right triangle: one side will be the distance to the screen and the other will be the separation of the bright fringe (which is what you're trying to find).
 
  • #19
No. Once you have the angle, set up a right triangle: one side will be the distance to the screen and the other will be the separation of the bright fringe (which is what you're trying to find).

i know that i have one angle which is 90 degrees and one which is 7,71 and then i know that the line to the m0 is 120 cm

then i think it is tan(7,76)=one side divided with the other side
distance = 120 cm -> 1,2 m*tan7,71 =0,162 m

if it isn't like this then i dont think i undestand the last thing u wrote
 
  • #20
Doc Al
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i know that i have one angle which is 90 degrees and one which is 7,71 and then i know that the line to the m0 is 120 cm

then i think it is tan(7,76)=one side divided with the other side
distance = 120 cm -> 1,2 m*tan7,71 =0,162 m
Looks good.
 
  • #21
Looks good.

Really :smile: u really dont know how happy i got right now all thanks to you.

so it is right that the distance is 0, 162 meters between the m0 and m1 blue ? you dont think the distance is to small at all ?
 
  • #22
by the way this webpage what is it presicely? a page to help people solve physics question or is it more than that ? and can anyone help or are u some kind of teacher since u do seem better than even teachers at school ? is it something from the UK or other country
 
  • #23
Doc Al
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Really :smile: u really dont know how happy i got right now all thanks to you.
Good!
so it is right that the distance is 0, 162 meters between the m0 and m1 blue ? you dont think the distance is to small at all ?
16 cm is not that small.
 
  • #24
Doc Al
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45,209
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by the way this webpage what is it presicely? a page to help people solve physics question or is it more than that ? and can anyone help or are u some kind of teacher since u do seem better than even teachers at school ? is it something from the UK or other country
We have members from all over the world.

One of the things we do is provide help to students. The only qualification for providing help is that you know what you're doing. Any one is welcome to jump in and provide help.

(Members who have proven themselves as expert and knowledgeable helpers may be awarded the HW Helper medal in recognition of their contribution to the site.)
 
  • #25
ok i just found the page today and i thought it was great that students can have such options for help since the teachers don't always have time.

at first i wasn't sure how to ask for help here since even though i do study english we dont learn how to talk about science in english so at first i was very afraid that no one would understand what i was asking about, but it went better than i thought and i felt like even though i know that i have written very strangely and sometimes didnt understand what you were saying and asked the same things a lot of times, you still took me serious and helped me.

so i guess that you have the HW award since you seem better than teachers at this stuff :=)
 

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