How Far is the Third-Order Bright Fringe from the Central Maximum?

[noobie!]

Homework Statement

Monochromatic light passes through 2 narrow slits 0.25mm apart and forms an interference pattern on screen 1.5m away.If light with a wavelength of 680nm is used,what is the distance between center of central maximum and center of the third-order bright fringes?

Homework Equations

sorry don't have..

The Attempt at a Solution

d=0.25mm ,y=1.5m lambda=680x10^-9, central max=0, third order=3
i tried using sin@=m x lambda /d ,but y is given..i think i should have use it but then i don't know where n when to use..can you please hint me?

 

[noobie!]

sorry,i've done it already,im so sorry..

 

[thomate1]

I would first like to commend you for attempting to solve this problem using the correct equation. Sin@=m x lambda/d is indeed the correct formula to use for calculating the distance between fringes in a double-slit interference pattern. However, you are correct in realizing that the given distance (y) is not being used in this equation.

To solve this problem, we must first convert the given distance between the slits (d) from millimeters to meters, as the wavelength (lambda) is given in meters. This gives us d = 0.00025m.

Next, we can use the given distance (y) of 1.5m to find the angle (@) between the central maximum and the third-order bright fringe. We can use the formula y = d x tan@ to solve for @, which gives us @ = 0.015 radians.

Now, we can plug in the values we have calculated into the equation sin@=m x lambda/d. We know that for the third-order bright fringe, m=3, so we can rearrange the equation to solve for the distance between the center of the central maximum and the third-order bright fringe (x). This gives us x = (3 x lambda x y)/d = 0.036m.

Therefore, the distance between the center of the central maximum and the third-order bright fringe is 0.036m, or 3.6cm. I hope this helps! Remember to always pay attention to the units and use the correct formula for the given problem.