In Young's double slit experiment, we consider two
electromagnetic waves having the same amplitudes. An interference
pattern consisting of bright and dark fringes is observed on the
screen. The distance between the slits is mm, the
wavelength for both waves is mm and the distance
from the aperture screen to the viewing screen
is mm.
In Young S Double Slit Experiment We Consider Two Electromagnetic Waves Having The Same Amplitudes An Interference Pat 1 (85.5 KiB) Viewed 29 times
Problem 2 [20 points) Fill in the empty boxes with your answer (YOUR ANSWER MUST BE ONLY A NUMBER; DO NOT WRITE UNITS; DO NOT WRITE LETTERS). Select the correct unit from the multiple choice drop down menu list next to your answer. In Young's double slit experiment, we consider two electromagnetic waves having the same amplitudes. An interference pattern consisting of bright and dark fringes is observed on the screen. The distance between the slits is 0.0036 m, the wavelength for both waves is 6.8.10-7 m and the distance from the aperture screen to the viewing screen is 9 m. a) [1 point] Which formula can be used to calculate the total irradiance resulting from the interference of the two waves? (refer to the formula sheet and select the number of the correct formula from the list) b) [5 points] The irradiance from one of the waves is equal to 265 W/m2. Using the correct equation from part a) find the location, y of the third maxima of total irradiance. y= c) [5 points) Find the location, y of the fifth minima of total irradiance. y = d) [1 point] The distance Ay between two consecutive maxima is given by: (6.8-10-)(0.0036) (6.8-10-7)(9) 0.0036 (9) (0.0036) 9 6.8-10-7 e) [3 points] Calculate Ay. Δy= f) [5 points] The location of the tenth maxima is located at y=0.017 m. Calculate its corresponding total irradiance (1 =6.8-10-7m; d = 0.0036 m; L=9 m; 10 = 265 W/m2). /=
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