Young's Experiment (destructive interference)

[roam]

Homework Statement

I'm having some trouble understanding the following solved problem:

In Young’s experiment, narrow double slits 0.20 mm apart diffract monochromatic light onto a screen 1.5 m away. The distance between the 5th minima on either side of the zeroth-order maximum is measured to be 34.73 mm. Determine the wavelength of the light.

Solution:

Position of mth minima relative to central maximum is:

$(m-\frac{1}{2}) \lambda = a \ sin \theta =\frac{ay_m}{L} \implies y_m = \frac{L(m-\frac{1}{2})\lambda}{a}$

Distance on the screen between the 5th minima on either side of the central maximum is:

$\Delta y = y_5-y_{-5} = \frac{(4.5-(-4.5))L \lambda}{a} = \frac{9L \lambda}{a}$

$\therefore \ \lambda = \frac{\Delta y a}{9L} = 514 \ nm$

I don't understand why they have used $(m-\frac{1}{2}) \lambda$ as the condition for destructive interference, instead of $(m+\frac{1}{2}) \lambda$?

Homework Equations

For destructive interference my textbook uses:

$d \ sin \theta_{dark} = (m+\frac{1}{2}) \lambda$

(this is an approximation)

The Attempt at a Solution

If I use -1/2, then 5th minima turn out to be are 9λ apart, just like the model answer.

However when I use +1/2 I get a totally different solution:

$\therefore \ \lambda = \frac{\Delta y a}{(5.5-(-5.5)) L}= \frac{\Delta y a}{11 L} =421 \ nm$

So, why do they use -1/2? And which method is correct?

Any help is greatly appreciated.

 

[STEMucator]

I think both you and your book might be wrong on this one. I'm getting $1029nm$ as the answer.

I used: $\frac{wx_n}{L} = (n - \frac{1}{2}) \lambda$

Where $n$ represents a nodal line number.

 

[roam]

I think both you and your book might be wrong on this one. I'm getting $1029nm$ as the answer.

I used: $\frac{wx_n}{L} = (n - \frac{1}{2}) \lambda$

Where $n$ represents a nodal line number.

No, that's wrong because you are not given y, you are given $\Delta y$.

 

[roam]

My textbook ("Optics" by Hecht) uses "m+1/2", while my other book ("Introduction to Optics" by Pedrotti) uses "m-1/2" for destructive interference. The two methods clearly produce different results!

Why in this particular situation, do we need "-1/2" and not "+1/2"?

 

[STEMucator]

No, that's wrong because you are not given y, you are given $\Delta y$.

No, you are not given $\Delta x$.

$\Delta x$ represents the distance between two consecutive nodal or antinodal fringes.

$x_n$ represents the distance from a minimum to the right bisector (central maximum).

The distance between the 5th minima on either side of the zeroth-order maximum is measured to be 34.73 mm

They are not talking about the distance between two consecutive node/antinodal fringes, rather the distance from the 5th minima on either side of the fringe pattern to the right bisector.

Also, because the fringe pattern is symmetric, I believe it wouldn't matter whether you used $(n - 1/2)$ or $(n + 1/2)$.

It depends on which direction the author likes to measure I believe.

 

[roam]

No, you are not given $\Delta x$.

$\Delta x$ represents the distance between two consecutive nodal or antinodal fringes.

$x_n$ represents the distance from a minimum to the right bisector (central maximum).

You are using a different notation. I never defined $\Delta y$ as the distance between consecutive fringes! It is the distance between the position of 5th minima on one side, to the 5th minima on the other.

$y$ itself is the linear positions of fringes measured along the screen from the bisector you mentioned.

 

[STEMucator]

You are using a different notation. I never defined $\Delta y$ as the distance between consecutive fringes! It is the distance between the position of 5th minima on one side, to the 5th minima on the other.

$y$ itself is the linear positions of fringes measured along the screen from the bisector you mentioned.

Indeed I use a different notation myself, but that really doesn't change the logic too much.

Just use $\frac{wy}{L} = (n - \frac{1}{2}) \lambda$ then.