Waves- Interference/Double-slit experiment

[Jodi]

Hi; Could someone please help me with the following question: In a double-slit experiment it is found that blue light of wavelength 460 nm gives a second-order maximum at a certain location on the screen. What wavelength of visible light would have a minimum at the same location? Can i use the equation dsin(theta) = m(lamda)? If I use this, i plug in the answer I get for dsin(theta) into this equation: dsin(theta) = (M+0.5)(lamda)? Is this right, because it doesn't work. Thanks for your help.

 

[Doc Al]

Looks good to me. Realize that there are many wavelengths of light that will have a minimum at that location, but not all will be visible. Plug in a few values of M and solve for $\lambda_2$ until you find one in the visible range.

 

[thepatientmental]

In a double-slit experiment, the interference pattern is created by the superposition of two waves from the two slits. This pattern is characterized by bright and dark fringes, with the bright fringes being the locations where the waves are in phase and the dark fringes being the locations where the waves are out of phase.

To find the wavelength of visible light that would create a minimum at the same location as the second-order maximum for blue light of wavelength 460 nm, we can use the equation dsin(theta) = m(lamda), where d is the distance between the two slits, theta is the angle between the line connecting the two slits and the location on the screen, m is the order of the interference and lambda is the wavelength of the light.

Since we are looking for a minimum at the same location as the second-order maximum, we can set m = 2 and plug in the values for d and lambda. This gives us:

dsin(theta) = 2(460 nm)

Now, to find the wavelength of the light that would create a minimum at this location, we can rearrange the equation to solve for lambda:

lambda = (dsin(theta))/2

Plugging in the values for d and sin(theta), we get:

lambda = (460 nm * sin(theta))/2

Since the value for sin(theta) will be the same for both the second-order maximum and the minimum at the same location, we can use the same value for it. This means that the wavelength of the light that would create a minimum at the same location as the second-order maximum for blue light of wavelength 460 nm would be:

lambda = (460 nm * sin(θ))/2

So, yes, you can use the equation dsin(theta) = m(lamda) to find the wavelength of the light that would create a minimum at the same location as the second-order maximum. However, make sure to use the correct values for d and sin(theta) to get the right answer. I hope this helps!