MHB Interference in a Double-Slit Experiment

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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.
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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.
 
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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?
 

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