Another Nature of Light Problem

[Mitchtwitchita]

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?

Homework Equations

Sin Theta = lambda/w

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 1.17:1


I'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?

 

[anathema]

Your approach to solving this problem is correct. The first dark fringe is observed at a smaller angle for slit 1, indicating a wider slit width. Your calculation for the ratio of the largest slit width to the smallest is also correct. Keep up the good work! If you have any further questions or difficulties, please don't hesitate to ask. Good luck with your studies!

 

[m2003]

Your approach to solving this problem seems correct. It's important to use the equation Sin Theta = lambda/w to determine the slit widths for each case. Your explanation for why slit 1 is wider is also correct. Keep practicing and you will become more comfortable with these types of problems. If you are still having difficulties, I suggest seeking additional help from your teacher or a tutor.