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?

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

radiation

explanation:

heat is transmitted by radiation

d. platinum would have the greatest temperature increase because it has the lowest specific heat.