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

Answer:

  λ = 467.3 nm and Δy = 2.9 10⁻³ m

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

Answer:

1)  λ = 467.3 10⁻⁹ m = 467.3 nm

2) Δy = 2.9 10⁻³ m

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