How to Find y in a Double Slit Experiment Without Intensity Values?

[zehkari]

Homework Statement

Homework Equations

(1) Φ = 2π*(dsinθ/λ)
(2) I_total = I₀ cos²(Φ/2)
(3) I_total = I₀ cos²(πdy/λR)

λ = 585 * 10^-9 m
R = 0.700 m
d = 0.320 * 10^-3 m
y = ?

The Attempt at a Solution

a) I used eqn (1) and divided the total oscillations by a complete oscillation to find the remainder. Which gave 4.59 rad. Which I think is right.

b) However, I am stuck on this question. I don't understand how to find y from not knowing intensity values. Can I work out the total itensity from eqn (2), now that I have the phase angle? Then find y from eqn (3)?

Any help or direction pointing would be appreciated. Somethings not clicking with this sub category of young's double slit.

 

[BvU]

Hi,

not knowing intensity values

You have as a given $I/I_0 = 1/2$, isn't that sufficient ?

 

[zehkari]

Ah, thank you yeah.

So,

Using eqn (3),

y = λ*R*cos^-1(sqrt (I/I₀)) / π*d

Which with I/I₀ = 1/2 , I get 18 mm for y.

If that looks alright, thanks again for your help BvU.

 

[TSny]

When taking the inverse cosine, should your calculator be in degree or radian mode?

 

[zehkari]

When taking the inverse cosine, should your calculator be in degree or radian mode?

Hey,

There are different values for degrees and radian. Radian gives 3.2 * 10^-4. Which is more realistic for distance between slits.
I never thought about radian or degree for inverse. I assumed either was ok. Could you explain why you have to use radian here for inverse cosine?

Many thanks.

 

[TSny]

Could you explain why you have to use radian here for inverse cosine?

Consider equation (1) in your first post. For the case where the path difference dsinθ equals one wavelength λ, the phase difference Φ should be a full cycle. That is, Φ would be 360 degrees or 2π radians. What does equation (1) give in this case? Note that (1) is used to get (3).

 

[zehkari]

From question a) and using eqn (1) I obtained an answer in degrees and then converted to radian. I am assuming then that Φ has to be in radian? Then as eqn (3) is derived from eqn (1), the calculator should be in radian?

Thanks again.

 

[TSny]

From question a) and using eqn (1) I obtained an answer in degrees and then converted to radian.

Equation (1), as written, gives the angle Φ in radians. If you want equation (1) to give Φ in degrees, you would have to rewrite it with the factor of 2π replaced by 360.