Single slit diffraction pattern

AI Thread Summary
A single slit diffraction pattern is analyzed with the first minimum at 40 degrees and a wavelength of 410 nm, while a second light source produces a second minimum at 60 degrees. The width of the slit is calculated using the formula w=2(λ)L/a, but initial attempts yielded incorrect results due to angle conversion errors. It is emphasized that the small angle approximation applies only when angles are in radians, which is crucial for accurate calculations. The discussion highlights the importance of correctly interpreting the problem and using appropriate formulas. The final calculations lead to a clearer understanding of the diffraction pattern and its parameters.
kevnm67
Messages
43
Reaction score
0

Homework Statement


A single slit forms a diffraction pattern, with the first minimum at an angle of 40 from the central maximum. Monochromatic light of 410 nm wavelength is used. The same slit, illuminated by a different monochromatic light source, produces a diffraction pattern with the second minimum at a 60 angle from the central maximum. (Remember to convert degrees to radians – Recall: 2πrad=〖360〗^0.)
A) What is the width of the slit?
a. 587 nm b. 103 nm c. 1170 nm

B) What is the wavelength of the second light source?
a. 587 nm b. 307 nm c. 615 nm

C) If the screen is placed 1 meter away, what is the width of the central maximum for the 410 nm light source?
a. 0.70 m b. 1.4 m c. 2.



Homework Equations



w=2(λ)L/a

The Attempt at a Solution



a) I solved for "a" and got 7 x 10-9 but don't know where to go from here?
 
Physics news on Phys.org
kevnm67 said:
a) I solved for "a" and got 7 x 10-9 but don't know where to go from here?
Show how you got that answer.

Did you copy the problem correctly? I don't see any of the answer choices as being correct.
 
Doc Al said:
Show how you got that answer.

Did you copy the problem correctly? I don't see any of the answer choices as being correct.

Yep, it's copied correctly...

for a I did:

theta = p (lamda/a)
and I get 10.25, not the other no., I used the wrong angle.
 
kevnm67 said:
Yep, it's copied correctly...

for a I did:

theta = p (lamda/a)
Ah... that formula is only an approximation. It's only true in the small angle limit, where sin(theta) ≈ theta, where theta is in radians. (I wondered why they mentioned radians!)

Using that approximation, I can see how they get their answers.
and I get 10.25, not the other no., I used the wrong angle.
Redo this calculation. In using the formula, make sure you convert the angle to radians.
 
Ok. I got it all now...radians! Thanks again, Dr. Al, ill be sure to give you credit on my final tomorrow ;)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top