course phys 202
1.The third-order fringe of 610 nm light created by two narrow slits is observed at 18 deg. How far apart are the slits?- Since it is third order, it has 3 wavelengths, so 3 * 610nm= 1830cm. Distance is added to this because of the slit spacing which uses the equation a sin (18degrees), then substitute a= 1830/sin 18 degress= 5920nm
2.460 nm light gives 2d-order max on screen; what wavelength would give a minimum?
-second order (2d) dsinthea= 2(460)= 920 nm since there are two lights we have to find the destructive interference from the other light
= dsinthea= (m+.5) 'lambda=(m+1/2)'lambda
920=(m+1/2)'lambda
m=0 920= (0+.5)= 1.84*10
m=1 920nm=(1+1/2)'lambda =613nm
m=2 920nm=(2+1/2)'lambda=368 nm
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