Youngs double slit exp.

[ibysaiyan]

Homework Statement

In a Youngs dbl. slit experiment using green light the fringe width was observed to be 0.20mm.if red light replaces green light,what will be the fringe width?

Homework Equations

hmm lambda=ay/d but (no info is given on rest)=/
wavelengths:green=5.2x10^-7m ,red=6.5x10^-7m

The Attempt at a Solution

well i don't know if its correct but i tried to solve it.by dividing the wavelength of of red over greens and i got 1.25m which i multiplied to (2x10^-4) <---- which is the fringe width of green .=/ i totally made it up .sos here =)

 

[The Bob]

Hey ibysaiyan,

Your method is right but now I want you to explain why it is. What do we know about d, n and L in the equation:

$$n \lambda \ = \ \frac{xd}{L}$$

in relation to this question?

The Bob

http://en.wikipedia.org/wiki/Double-slit_experiment

 

[nlightner]

I would like to clarify that the fringe width in a Young's double slit experiment is dependent on the wavelength of the light used, as well as the distance between the slits and the screen. The equation you mentioned, lambda=ay/d, is the formula for calculating the wavelength of light using the fringe width (a), the distance between the slits (d), and the distance between the slits and the screen (y). However, this equation assumes that the light used is monochromatic (single wavelength).

In this scenario, the fringe width for green light is 0.20mm and the wavelength is 5.2x10^-7m. If we were to replace the green light with red light, which has a longer wavelength of 6.5x10^-7m, the fringe width would also change. Using the same equation, we can calculate that the new fringe width for red light would be 0.25mm. This is because the longer wavelength will result in a wider spacing between the fringes.

In conclusion, the fringe width in a Young's double slit experiment is directly proportional to the wavelength of light used. Therefore, replacing the green light with red light will result in a wider fringe width of 0.25mm.