physics Problems. Optics and such

Question

The wavelength of an EM wave is 650 nm when the wave is in a medium with index of refraction of 1.51. What is the frequency of the EM wave?

A light with wavelength 190 nm is incident on a double slit and a fringe pattern is observed on a screen 60.0 cm away. The distance between the bright fringes on either side of the centerline is 29.2 cm.

What is the slit separation distance

A double slit has a red LASER operating at 781 nm incident on it and a fringe pattern is observed on a distant screen (small angle approximation is valid). The fringe spacing is 3.4 mm. Note that the spacing between the fringes is the distance between the centerline and the first maximum (bright). 

The source is changed to a new LASER with wavelength 450 nm (set-up otherwise remains the same). How far away will the 3rd minimum (dark) be from the centerline?



Answer 1

Q1. Remember that EM waves traveling in a pure vacuum travel at the speed of light (c), and this speed changes once the waves encounter a different medium. The index of refraction stated is what describes this change in speed according to the equation n = c/v where n is the index of refraction, c is the speed of light, and v is the speed of the wave in the medium. With this and the trusty old λ=v/ƒ equation you should be able to discover the frequency of the wave.

Q2. What causes the interference pattern on the other side of a double slit is the constructive and destructive interference of the light waves traveling from each of the slits. This is due to the difference in path length from each of the slits to the screen. Given that the distance between slits is much less than the distance to the screen, we can use the equation d*sinθ = m*λ to find the angle theta (θ) between the line connecting the light fringes on the screen to the middle point between the two slits on the close wall and the horizontal(line perpendicular to the screen). d is the slit distance, m is an integer, and λ is the lights wavelength. Each m value corresponds to each bright fringe on the wall such that m=1 is the first bright fringes outside the centerline, m=2 are the next because of the constructive interference that occurs between the light rays as the path length increases or decreases by an integer value of wavelength. I'll let you decide what m value you should use for this problem and solve for d in terms of theta. With this you should be able to use trigonometry paired with the knowledge of how far back the screen is to solve for the angle theta that gives the separation distance in the problem and thus the slit distance d using the equation you manipulated above.

Q3. Take a shot at this one on your own again using the d*sinθ = m*λ equation and remember for bright fringes you have to use integer values such that you get constructive interference.