Double-Slit sin (2 theta) help?

[MadMatt]

I'm in physics I and only in Trig for the first time this year (Jr in HS).
I do not remember sin (2 theta) in any previous lessons.
I need in help in the following problem using it.

In a double-slit experiment with monochromatic light and a screen at a distance of 1.50 , the angle between the second-order bright fringe and the central maximum is 0.0230 rad. If the separation distance of the slits is 0.0350 mm, what are (a) the wavelength and color of the light and (b) the lateral displacement of the fringe.

I am so confused with the question. I'm not even sure i need the sin (2 theta) in here. Any help will be appreciated.

I am new to the site (found in a google search). So if this is the wrong forum, I appologize.

 

[HallsofIvy]

I'm moving this to the physics homework forum.

 

[kyle8921]

Hello,

Thank you for reaching out for help with your physics problem. I am happy to assist you.

First, let's talk about the double-slit experiment. In this experiment, a beam of light is shone through two parallel slits and the resulting pattern on a screen is observed. This pattern is known as an interference pattern and is caused by the overlapping of the waves from the two slits.

Now, let's address the use of sin (2 theta) in this problem. In this experiment, theta represents the angle between the central maximum (the brightest spot on the screen) and the first-order bright fringe (the first bright spot next to the central maximum). The formula for the position of the bright fringes is given by d sin (theta) = mλ, where d is the distance between the slits, m is the order of the fringe, and λ is the wavelength of the light.

In your problem, the angle between the second-order bright fringe and the central maximum is given as 0.0230 rad. We can use this information to find the wavelength of the light. Rearranging the formula, we get λ = d sin (theta) / m. Plugging in the values, we get λ = (0.0350 mm)(sin 0.0230 rad) / 2 = 5.75 x 10^-7 m.

Now, to determine the color of the light, we can use the relationship between wavelength and color. The visible spectrum ranges from approximately 400 nm (violet) to 700 nm (red). Since our calculated wavelength is 5.75 x 10^-7 m, it falls within the visible range and can be classified as green light.

Moving on to part (b) of the problem, we need to find the lateral displacement of the fringe. This is the distance between the central maximum and the second-order bright fringe. We can use the formula d sin (theta) = mλ again, but this time we are solving for d. Rearranging the formula, we get d = mλ / sin (theta). Plugging in the values, we get d = (2)(5.75 x 10^-7 m) / sin 0.0230 rad = 0.0500 mm.

I hope this explanation helps you understand the problem better. Remember, in physics it is important to understand the concepts and formulas, rather than just memorizing