explain A Young\'s interference experiment is performed with blue-green laser li

Question

explain

A Young's interference experiment is performed with blue-green laser light. The separation between the slits is 0.soo mm. and the screen is located 3.10 m from the slits. The first bright fringe is located 3.12 mm from the center of the interference pattern. What is the wavelength of the laser light? Need Help? Reeda -2 points serPSEs 37 p M FB My Notes o Ask Your Teacher A pair of narrow, parallel s separated by o.220 mm is illuminated by green light (a s46.1 nm). The interference pattern is observed on a screen 1.10 m away from the plane of the parallel slits. (a) Calculate the distance from the central maximum to the first bright region on either side of the central (b) calculate the distance between the first and second dark bands in the interference pattem. Need Help?

Explanation / Answer

m = d sin

& sin = y/L for m = 1

d = 0.5mm

y = 3.12 mm

L = 3.1 m

sin = 3.12 *10^-3/ 3.1 = 1.006 * 10^-3

= d*sin = 0.5*10^-3* 1.006* 10^-3

= 5.03*10^-7 = 503 nm

B1. by above equation y = m* *L/d = 1*546.1*10^-9*1.1/(0.22*10^-3) = 2.73*10^-3 m

B2. for first dark band y1 = y/2 = 1.365*10^-3 m and for second dark band y2 = 1.5*2.73*10^-3 = 4.095*10^-3 m

difference between bands = dy = y2 - y1 = (4.095 - 1.365)*10^-3 = 2.73*10^-3 m = 2.73 mm

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