Question

A 680 nm laser illuminates a double slit apparatus with a slit separation distance of 7.83 μm. On the viewing screen, you measure the distance from the central bright fringe to the 2nd bright fringe to be 88.2 cm. How far away (in meters) is the viewing screen from the double slits?

Answer 1

Answer:

L = 5.08 10⁻¹ m = 50.8 cm

Explanation:

In a double slit experiment for constructive interference is given by

           d sin θ = m λ

let's use trigonometry to find a relationship with the distance

           tan θ = y / L

these experiments are very small angles

           tan θ = $\frac{sin \ \theta}{cos \ \theta}$ = sin θ

when substituting

           sin θ = y / L

substituting in the first equation

           d y / L = m λ

           

           L = $\frac{d \ y}{m \ \lambda}$

let's calculate

           L = $\frac{ 0.783 \ 10^{-6} \ 88.2 \ 10^{-2} }{2 \ 680 \ 10^{-9}}$

           L = 5.08 10⁻¹ m

Answer 2

The distance of viewing screen from the double slits will be L = 5.08 *10⁻¹ m = 50.8 cm

What is constructive interference?

This position, where the resulting wave is larger than either of the two original, is called constructive interference.

In a double slit experiment for constructive interference is given by

d sin θ = m λ

let's use trigonometry to find a relationship with the distance

$tan\theta=\dfrac {y}{L}$

these experiments are very small angles

tan θ =  = sin θ

when substituting

$sin\theta = \dfrac{y}{l}$  

substituting in the first equation

 

$\dfrac{dy}{L}=m\lambda$

$L=\dfrac{dy}{L\lambda}$

let's calculate

$L=\frac{0.783\times 10^{-6}\times 88.2\times 10^{-2}}{2.680\times 10^{-9}}$

L = 5.08 10⁻¹ m

Hence the distance of viewing screen from the double slits will be L = 5.08 10⁻¹ m = 50.8 cm

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