Diffraction grating problem involving Snell's Law

[Sarah Hallsway]

Homework Statement

A 600 line/mm diffraction grating is in an empty aquarium tank. The index of refraction of the glass walls is [n][/glass] = 1.50. A helium-neon laser (lambda=633 nm) is outside the aquarium. The laser beam passes through the glass wall and illuminates the diffraction grating.

a. What is the first order diffraction of the laser beam?
b. What is the first-order diffraction angle of the laser beam after the aquarium is filled with water? ([n][/water] = 1.33)

Homework Equations

Diffraction equation= m(lambda)=d(sin(theta1)+sin(theta2))
where m= diffraction order
d= grating spacing
theta 1= incident angle
theta 2= diffraction angle

and possibly Snell's law,
n1sin(theta1)=n2sin(theta2)

The Attempt at a Solution

My first try at a gave me the answer -41.80 degrees as the angle of diffraction. I assumed that the incoming angle (from the laser to the glass of the aquarium) was 90, then used snell's law to calculate the angle at which the laser bent, which was 41.81 degrees. I used 41.81 degrees in the diffraction gradient equation I provided above and got my theta2 to be equal to -41.80 degrees.

I know I probably went wrong assuming that the incoming angle was 90, and I also noticed that theta1 and theta2 were very similar which is not usually the case for diffraction gradient problems.

If someone could even just point me in the right direction, I would be very appreciative. For reference, the correct answers to this problem are a. 23.3 degrees and b. 16.6 degrees. Thank you in advance!

 

[TSny]

Welcome to PF!

I assumed that the incoming angle (from the laser to the glass of the aquarium) was 90, then used snell's law to calculate the angle at which the laser bent, which was 41.81 degrees.

If the laser light hits the glass of the aquarium perpendicularly to the glass, what is the angle of incidence of the laser light that you would use in Snell's law?

 

[Sarah Hallsway]

Welcome to PF!

If the laser light hits the glass of the aquarium perpendicularly to the glass, what is the angle of incidence of the laser light that you would use in Snell's law?

Oh, it would be 0, right? When I use 0 for theta0 in snell's law, I get that theta1 also equals 0, and when I put that into the equation m(lambda)=d(sin(theta1)+sin(theta2)), I get an extremely small value for the first order diffraction angle (theta2). Where am I going wrong?

Also, is the diffraction grating on the bottom of the tank or on the sides? In other words, is the laser pointer parallel to the diffraction grating or perpendicular?

 

[TSny]

Oh, it would be 0, right?

Yes.

When I use 0 for theta0 in snell's law, I get that theta1 also equals 0,

I'm not sure what angle theta1 refers to.

and when I put that into the equation m(lambda)=d(sin(theta1)+sin(theta2)), I get an extremely small value for the first order diffraction angle (theta2). Where am I going wrong?

Can you explain the meaning of theta1 and theta2?

Also, is the diffraction grating on the bottom of the tank or on the sides? In other words, is the laser pointer parallel to the diffraction grating or perpendicular?

I would assume that the laser beam hits the grating at right angles to the plane of the grating.