MHB Interference in a Double-Slit Experiment

Click For Summary
In the double-slit experiment, the distance between the slits is 5.0 mm, and they are positioned 1.0 m from the screen. The interference patterns are created by light wavelengths of 480 nm and 600 nm. To find the separation between the third-order bright fringes (m=3) for both wavelengths, the equation d*sin(θ) = m*λ is used. The position on the screen can be determined by the relationship OP = D*sin(θ). The discussion emphasizes the need to calculate θ to find the specific positions of the bright fringes.
annikaw
Messages
1
Reaction score
0
In a double-slit experiment, the distance between slits is 5.0mm and the slits are 1.0m from the screen. Two interference patterns can be seen on the screen: one due to light of wavelength 480nm, and the other due to light of wavelength 600nm. What is the separation on the screen between the third-order (m=3) bright fringes of the two interference patterns?

From my notes, I find that $d\sin\theta=m\lambda$.

However, I do not know how to get values of $\theta$ that can be put to the equation to find the difference in separation.
 
Last edited by a moderator:
Mathematics news on Phys.org
annikaw said:
In a double-slit experiment, the distance between slits is 5.0mm and the slits are 1.0m from the screen. Two interference patterns can be seen on the screen: one due to light of wavelength 480nm, and the other due to light of wavelength 600nm. What is the separation on the screen between the third-order (m=3) bright fringes of the two interference patterns?

From my notes, I find that $d\sin\theta=m\lambda$.

However, I do not know how to get values of $\theta$ that can be put to the equation to find the difference in separation.

Hi annikaw! Welcome to MHB! (Smile)

Let's get a drawing of the situation:
View attachment 4506

We get a bright fringe when $d\sin\theta=m\lambda$.
The corresponding position on the screen is $OP=D\sin\theta$.

Can you deduce where the 2 requested fringes are?
 

Attachments

  • phys6_1f_11.png
    phys6_1f_11.png
    13.6 KB · Views: 97
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

Similar threads

  • · Replies 15 ·
Replies
15
Views
1K
Replies
23
Views
2K
  • · Replies 15 ·
Replies
15
Views
1K
Replies
55
Views
5K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 10 ·
Replies
10
Views
653
Replies
32
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K