How Does Immersion in a Liquid Affect Diffraction Fringe Spacing?

[jasper10]

Homework Statement

A monochromatic light source is incident normally on a diffraction grating with 600 lines per mm. The beam passes through the grating and the fringes are observed on a screen 5m away. The apparatus is then immersed in a liquid whose refractive index is 1.3. In consequence the distance between the adjacent bright fringes is reduced by 0.45m. Find the wavelength of this light source in air.

Homework Equations

The Attempt at a Solution

I derived the formula: x = n\lambdaD / d , where D = 5m and d = 1 / (600 x 10^3) = 1.67 x 10^-6m

then I said: x - 0.45 = (1.3\lambda x D x n)/ d

but there are two unknowns, and this is where I'm stuck. I have the feeling that simultaneous equations will be needed.

Thanks for any help!

 

[jasper10]

How can I use the refractive index to solve the question?

 

[Doc Al]

I derived the formula: x = n\lambdaD / d , where D = 5m and d = 1 / (600 x 10^3) = 1.67 x 10^-6m

then I said: x - 0.45 = (1.3\lambda x D x n)/ d

Two problems:
(1) The distance between adjacent fringes is not reduced by 0.45 m, it is reduced to 0.45 m. In other words: x = 0.45 m.
(2) In that formula, n is the order of the fringe, not the index of refraction.

How can I use the refractive index to solve the question?

How does the presence of the liquid affect the wavelength of the light?

 

[jasper10]

Two problems:
(1) The distance between adjacent fringes is not reduced by 0.45 m, it is reduced to 0.45 m. In other words: x = 0.45 m.

The question clearly says BY 0.45 m.

(2) In that formula, n is the order of the fringe, not the index of refraction.

Yes I'm aware of this: I used n to be 1 (first order maximum). Hence I am considering the angle between n=1 and n=0.

How does the presence of the liquid affect the wavelength of the light?

The liquid decreases the wavelength. does this mean: 1.3\lambda = 1.0\lambda' (assuming that the refractive index for air is n=1)

Can you please help me find a method of how to tackle this question? I still don't see it unfortunately! Thanks!

 

[Doc Al]

The question clearly says BY 0.45 m.

Hmmm... Did you edit your post? I don't recall reading that. In any case, OK.

Yes I'm aware of this: I used n to be 1 (first order maximum). Hence I am considering the angle between n=1 and n=0.

OK.

The liquid decreases the wavelength. does this mean: 1.3\lambda = 1.0\lambda' (assuming that the refractive index for air is n=1)

Yes. λ_air = 1.3λ_liquid

Can you please help me find a method of how to tackle this question? I still don't see it unfortunately!

Your initial idea of simultaneous equations was correct. Set up two diffraction equations, one for liquid and one for air.

 

[jasper10]

ok but which equation should I use?

\lambda

x - 0.45 = (1.3\lambda x D x n)/ d

and x = (\lambda x D x n) / d ?

Thanks for your help!

 

[Doc Al]

x - 0.45 = (1.3\lambda x D x n)/ d

Why 1.3λ? You want x_liquid = λ_liquid (D/d). (Expressed in terms of x and λ, of course.)

and x = (\lambda x D x n) / d ?

OK. (That's the equation for air.)

 

[jasper10]

Why 1.3λ? You want x_liquid = λ_liquid (D/d). (Expressed in terms of x and λ, of course.)

But I assume you want to express \lambda_liquid in terms of λ_air.

So i came to the conclusion:

In water: x - 0.45 = (Dλ_air/1.3) / d
In air: x = (λ_airD) / d
(I have not included n in these to equations, as I have taken n = 1)

Solving this gives: λ_air= 6.5 x 10^-7 m
Is this correct? I have a feeling that it isn't (because the question is worth a lot of marks- which hints that it takes more than a few steps!) :S Thanks!

 

[Doc Al]

Solving this gives: λ_air= 6.5 x 10^-7 m
Is this correct?

Looks good to me.