d = 190×10^-6m
Angle theta = 0.63°
n = 3
Since d Sin(theta) = n(lemda)
=) Lemda = 190×10^-6+ Sin(0.63) / 3 = 0.6964 ×10^-6 m
=) Wavelength lemda = 696.4×10^-9 = 696.4nm
(5%) Problem 2: Suppose light falls on double slits separated by 190 um. Randomized Variables d= 190 um 0= 0.63° What i...
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Suppose a double-slit interference pattern has its third minimum at an angle of 0.279° with slits that are separated by 287 μm. Class Management | Help HW 9 Begin Date: 7/27/2020 12:01:00 AM -- Due Date: 7/30/2020 11:59:00 PM End Date: 8/8/2020 11:59:00 PM (11%) Problem 5: Suppose a double-slit interference pattern has its third minimum at an angle of 0.279° with slits that are separated by 287 um. Randomized Variables A = 0.2790 d=287 um >> * Calculate the...
Suppose a double-slit interference pattern has its third minimum at an angle of 0.279° with slits that are separated by 287 μm. Class Management | Help HW 9 Begin Date: 7/27/2020 12:01:00 AM -- Due Date: 7/30/2020 11:59:00 PM End Date: 8/8/2020 11:59:00 PM (11%) Problem 5: Suppose a double-slit interference pattern has its third minimum at an angle of 0.279° with slits that are separated by 287 um. Randomized Variables A = 0.2790 d=287 um >> * Calculate the...