In a classroom demonstration, a beam of coherent light of wavelength 550 nm is i
ID: 1987816 • Letter: I
Question
In a classroom demonstration, a beam of coherent light of wavelength 550 nm is incident perpendicularly onto a pair of slits. Each slit has a width w of 1.2 x 10^-6 m, and the distance d between the centers of the slits is 1.8 x 10^-5 m from the slits. Your notebook shows the following setup for the demonstration.a. Calculate the frequency of the light.
b. Calculate the distance between tow adjacent dark fringes on the screen.
The entire apparatus is now immersed in a transparent fluid having index of refraction 1.4
C. what is the frequency of the light in the transparent fluid?
d. Does the distance between the dark fringes increase, decrease, or remain same? Explain your reasoning.
Explanation / Answer
a. f = v/lambda speed of light = 3x10^8 m/s f = (3x10^8 m/s)/(550x10^-9 m) f = 5.4546 x 10^14 Hz b. formula: x = (lambda* (m + 0.5) * l)/d since you lookin for distance between adjacent dark fringes: change in x = (lambda * change in m * l)/ d change in m = 1 because adjacent You did not give a distance between the slits and the screen. So, I did a quick google search and assumed you are doing an old AP physics problem. If not, substitute the 2.2m I used with whatever your distance to the screen is. change in x = (550*10^-9 m * 1 * 2.2 m)/ (1.8 x 10^-5 m) change in x = 6.72 cm c. Frequency will not change d. The distance will decrease. The change in medium does not affect the frequency, it affects the wavelength by causing it to decrease. Since wavelength is decreasing, the constructive and deconstructive interference points become closer together. Another way to look at it is the formula used in part b was dependent on wavelength and if wavelength decreases, then so must the distance between adjacent dark spots.