Double-Slit Calculation

1. May 11, 2014

vanceEE

1. The problem statement, all variables and given/known data
In Fig. 6.1 the distance from the two slits to the screen is 1.8 m. The distance CP is
2.3 mm and the distance between the slits is 0.25 mm.
Calculate the wavelength of the light provided by the laser.

2. Relevant equations
$$asinθ = nλ$$
$$w = \frac{λD}{a}$$

3. The attempt at a solution
$$θ = tan^{-1}(\frac{2.3×10^{-3}}{1.8}) = 0.0732°$$
Since P is first order, n = 1.
$$λ = (0.25×10^-3)(sin(0.0732°)) = 319.4 nm$$

According to the mark scheme it is incorrect:
$$λ = ax / D$$
$$= \frac{2 × 2.3 × 10^{–3} × 0.25 ×10^{–3}}{1.8}$$
$$= 639 nm$$

Why must I multiply my wavelength by two?

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Last edited: May 11, 2014
2. May 11, 2014

rpthomps

It's because that equation is for B to C (according to your diagram) not C to P. Or, in other words, bright fringe to bright fringe. Since, the dark fringe is halfway between the bright fringes, you get half the result.

3. May 11, 2014

vanceEE

$$θ = tan^{-1}(\frac{|CB|}{1.8}) = tan^{-1}(\frac{4.6*10^{-3}}{1.8}) = 0.1464°$$
B is 2nd order, hence n = 2.
$$(0.25*10^{-3})sin(0.1464°) = 2λ$$
$$λ = \frac{(0.25*10^{-3})sin(0.1464°)}{2} = 319.4 nm$$

Please explain; I am still getting the same wavelength.

4. May 11, 2014

vanceEE

Are the bright fringe to bright fringe considered to be 1st order? For example, would the middle-fringe (green) be zero and the first two greens on the top and bottom of the middle fringe be considered first order?

If so, then the equation would in fact be:
$$(0.25*10^{-3})sin(0.1464°) = λ$$
= 639 nm
Right?

Last edited by a moderator: May 6, 2017
5. May 11, 2014

rpthomps

It looks first order to me(1st bright fringe to the center). I would use the equation:

$$λ=\frac{mx_md}{L}$$ where

x_m is the distance from the center to the first bright fringe
m is the order (in this case 1)

6. May 11, 2014

rpthomps

Yes. You were thinking bright fringe to dark fringe is one order but in fact it is bright to bright. This is why you are getting half of what you should have got.

7. May 11, 2014

vanceEE

Ok, thanks for the help!