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alicia113
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1. Homework Statement
Two different single slits are used in an experiment involving one source of monochromatic light. With slit 1 in place, the first dark fringe is observed at an angle of 45 degrees. With slit 2, the first dark fringe is observed at an angle of 55 degrees.
a)Which slit is widest? Why?
b)What is the ratio of the largest slit width to smallest?2. Homework Equations
Sin Theta = lambda/w3. The Attempt at a Solution
a) Slit 1 would be the widest slit. This is because the first dark fringe is observed at less of an angle than slit 2. Therefore, the number of pairs of rays that destructively interfere are increased which causes the maxima to decrease.
b)Let Lambda = 720 x 10 e-9 m for both cases
Given: Theta(1) = 45 degrees
Theta(2) = 55 degrees
Slit 1: Sin Theta = Lambda/w
Therefore, Sin 45 degrees = 720 x 10 e-9 m/w
w = 720 x 10 e-9 m/Sin 45 degrees
=1.03 x 10 e-6
Slit 2: Sin Theta = Lambda/w
Therefore, Sin 55 degrees = 720 x 10 e-9 m/w
w = 720 x 10 e-9 m/Sin 55 degrees
=8.8 x 10 e-7 m
For Ratio: 1.03 x 10 e-6 m/8.8 x 10 e-7 m
=1.17
Therefore, the ratio of the largest slit width to the smallest is approximately 1.17:1I've been having some difficulties with these Nature of light problems and I was wondering if I'm going about them the right way. Can anybody help me?...Please?
Two different single slits are used in an experiment involving one source of monochromatic light. With slit 1 in place, the first dark fringe is observed at an angle of 45 degrees. With slit 2, the first dark fringe is observed at an angle of 55 degrees.
a)Which slit is widest? Why?
b)What is the ratio of the largest slit width to smallest?2. Homework Equations
Sin Theta = lambda/w3. The Attempt at a Solution
a) Slit 1 would be the widest slit. This is because the first dark fringe is observed at less of an angle than slit 2. Therefore, the number of pairs of rays that destructively interfere are increased which causes the maxima to decrease.
b)Let Lambda = 720 x 10 e-9 m for both cases
Given: Theta(1) = 45 degrees
Theta(2) = 55 degrees
Slit 1: Sin Theta = Lambda/w
Therefore, Sin 45 degrees = 720 x 10 e-9 m/w
w = 720 x 10 e-9 m/Sin 45 degrees
=1.03 x 10 e-6
Slit 2: Sin Theta = Lambda/w
Therefore, Sin 55 degrees = 720 x 10 e-9 m/w
w = 720 x 10 e-9 m/Sin 55 degrees
=8.8 x 10 e-7 m
For Ratio: 1.03 x 10 e-6 m/8.8 x 10 e-7 m
=1.17
Therefore, the ratio of the largest slit width to the smallest is approximately 1.17:1I've been having some difficulties with these Nature of light problems and I was wondering if I'm going about them the right way. Can anybody help me?...Please?