Question

1. A double-slit experiment is set up using red light (l = 701 nm). A first order bright fringe is seen ata given location on a screen. What wavelength of visible light (between 380 nm and 750 nm) would produce a dark fringe at the identical location on the screen?2. A new experiment is created with the screen at a distance of 2 m from the slits (with spacing 0.1 mm). What is the distance between the second order bright fringe of light with λ = 692 nm and the third order bright fringe of light with λ = 413 nm?

Answer 1

λ = 467.3 nm and Δy = 2.9 10⁻³ m

Step-by-step explanation:

The equation is d sin θ = m λ

For red, wavelength 'λ'= 701 mn = 701 x 10⁻⁹ m

and as it's a first-order interference so m = 1,

d sin θ = m λ (m=1)

d sin θ = λ₁ ---> (1)

In order to find the wavelength, the destructive interference occurs at this same point is given by

d sin θ = (m + ½) λ --->(2) ( m = 1 again for this case)

Substituting 'd sin θ' from eq(1) in above eq

λ₁ = (1 + ½) λ

λ = λ₁ /1.5

λ = 701 10⁻⁹ /1.5

λ = 467.3 nm

2) In order to find the distance to each strip, we'll use tangent function i.e

tan θ = y / x

By approximating tangent to the sine because angles are very small, we will have

tan θ = sin θ / cos θ = sin θ = y / x

We substitute

d y / x = m λ

m is 2 for this case

y₁ = m λ x / d

y₁ = 2 x 692x10⁻⁹ x 2 / 0.1x10⁻³

y₁ = 2.768 10⁻² m

the interference is considered third order m = 3 for the other wavelength

y₂ = 3 413 10⁻⁹ 2 / 0.1 10⁻³

y₂ = 2.478 10⁻² m

The two stripes will have a difference of

Δy = y₁ - y₂=> (2.768x10⁻² - 2.478x10⁻²)

Δy = 2.9 10⁻³ m

Answer 2

1) λ = 467.3 10⁻⁹ m = 467.3 nm

2) Δy = 2.9 10⁻³ m

Step-by-step explanation:

1) The interference experiment is described by the expression

.d sin θ = m λ

It indicates that the wavelength is λ= 701 mn = 701 10⁻⁹ m and we have a first-order interference whereby m = 1, let's find the angle for which it occurs

d sin θ = m λ

d sin σ = λ₁ (1)

We are asked to find the wavelength so that destructive interference occurs at this same point, which is described by

d sin θ = (m + ½) λ (2)

In this case also m = 1

We substitute 1 in 2

λ₁ = (1 + ½) λ

λ = λ₁ /1.5

Let's calculate

λ = 701 10⁻⁹ /1.5

λ = 467.3 10⁻⁹ m = 467.3 nm

2) In this new experiment we must find the distance to each strip

Let's use trigonometry

tan θ = y / x

Furthermore, the angles in these experiments are very small, so we can approximate the tangent to the sine

tan θ = sin θ / cos θ = sin θ = y / x

We substitute

d y / x = m λ

In this case m = 2

y₁ = m λ x / d

y₁ = 2 692 10⁻⁹ 2 / 0.1 10⁻³

y₁ = 2.768 10⁻² m

For the other wavelength the interference is the third order m = 3

y₂ = 3 413 10⁻⁹ 2 / 0.1 10⁻³

y₂ = 2.478 10⁻² m

The distance between these two stripes is

Δy = y₁ - y₂

Δy = (2,768 - 2,478) 10⁻²

Δy = 2.9 10⁻³ m