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IN YDSE for wavelength=589nm, the interference fringes have angular seprationof 3.50×10-3 rad. For what wavelength would the angular seprationbe10.0%greater?

Abhilasha jha , 8 Years ago
Grade 12
anser 1 Answers
Vikas TU
Expanded Angular division = 3.50 x 10^-4 
D1/D2 = 589/x 
3.50 x 10^-3/3.50 x 10^-4 = 589 x 10^-9/x 
=> 10 = 589 x10^-9/x 
=> x = 589 x10^-9/10 
=> x = 589 x 10^-10 m 
Last Activity: 8 Years ago
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