In a Young\'s double slit experiment, two narrow parallel slits are illuminated
ID: 1632000 • Letter: I
Question
In a Young's double slit experiment, two narrow parallel slits are illuminated by light of wavelength 557 nm. The slit separation is 0.180 mm. An observing screen is located 1.30 m from the slits. Calculate the distance between the first and second maxima of the interference pattern observed on the screen. NOTE: It is best to keep many significant figures during the calculation, and then round to THREE significant figures for your answer. m
A double-slit experiment is performed with light of wavelength 600 nm. The bright interference fringes are spaced 1.75 mm apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 430 nm? mm
The figure below shows the light intensity on a screen behind a double slit. The distance y is 2.0 cm. The slit spacing is 0.20 mm and the wavelength of the light is 640 nm. What is the distance from the slits to the screen? cm. The picture for this question just shows a graph with intensity on the y asix and "y" on the x axis. The max value on the y axis is 12 w/m^2.
Explanation / Answer
Doing first question
We know that any mth order bright fringe distance is given as
y=m*wavelength*D/d
where D is screen distance and d is slit width
putting m=2 and m=1 and subtracting we got distance between them
=wavelength*D/d=4.02*10^(-3) m