Diffraction & Interference problems

[mustang]

Problem 15.
By attaching a diffrection-grating spectroscope to an astronomical telescope, one can measure the spectral lines from a star and determine the star's chemicalcomposition. Assume the grating has 3224 slits/cm. The wavelength of the star's light is wavelength 1 = 463.200nm.
find the angle at which the second-order spectral line for wavelength 1 occurs. Answer in units of degrees.
How do you start?

 

[Doc Al]

How do you start?

Hmmm... you asked a similar question recently. The answer hasn't changed.
https://www.physicsforums.com/showthread.php?t=21348

 

[mustang]

Regards on problems

5.
Light falls on two slits spaced 0.374mm apart. The angle between the first dark fringe and the central maximum is 0.0589 degrees.
What is the wavelength of the light? Answer in units of nm.
Note: This is what i have done
d(sin thetha)=(m+1/2) * wavelength(or w)
(0.374e-3)(sin 0.0589)=(1+1/2)*w
256=w
is this right?

Problem 6. A sodium-vapor street lamp produces light that is nearly monochromatic. If the light shines on a wooden door in which there are two straight, parallel cracks, an interfernece pattern will form on a distant wall behind the door. The slits have a separation of 0.3740mm, and the second-order maximum occurs at an angle of 0.18046 degrees from the central maximum.
determine the wavelength of the light.
This is what I have done:
d(sin thetha)=(m+1/2) * wavelength(or w)
(0.3740e-3)(sin0.18046)=(2+1/2)*w
471 = w
Is this right?

 

[Doc Al]

Problem 5. ...
Note: This is what i have done
d(sin thetha)=(m+1/2) * wavelength(or w)
(0.374e-3)(sin 0.0589)=(1+1/2)*w
256=w
is this right?

No. The first dark fringe occurs for a path difference of 1/2λ, not (1 + 1/2)λ

Problem 6. ...
This is what I have done:
d(sin thetha)=(m+1/2) * wavelength(or w)
(0.3740e-3)(sin0.18046)=(2+1/2)*w
471 = w
Is this right?

No, same kind of problem. The second order maxima occurs for a path difference of 2λ.

You need to understand that the diffraction pattern goes like this: central max - dark fringe - 1st order maxima- dark fringe - 2nd order maxima... etc. Each maxima corresponds to another integral wavelength of path difference.

 

[mustang]

So for problem 5.

For problem 5. w=wavelength
(d)sin(thetha)=(m+1/2)*w
(0.374*10^-3)sin(0.0589)=1/2w
922 = w

 

[Doc Al]

For problem 5. w=wavelength
(d)sin(thetha)=(m+1/2)*w

I presume you mean: d sinθ = 1/2λ

(0.374*10^-3)sin(0.0589)=1/2w
922 = w

Check your arithmetic.

 

[mustang]

Whoops!

I rechecked my math and I found that w= does not equal 922 but 769.