Question

Sunlight is used in a double-slit interference experiment. The fourth-order maximum for a wavelength of 460 nm occurs at an angle of θ = 90°. Thus, it is on the verge of being eliminated from the pattern because θ cannot exceed 90° in Eq. 35-14. (a) What least wavelength in the visible range (400 nm to 700 nm) are not present in the third-order maxima? To eliminate all of the visible light in the fourth-order maximum, (b) should the slit separation be increased or decreased and (c) what least change in separation is needed?

Answer 1

Answer:

a)λ  = 6.133 10⁻⁷ m = 613.3 nm, b) d = 1.60 10⁻⁶ m , c) Δd = 0.24 10⁻⁶ m

Explanation:

The expression that describes the interference phenomenon for the double slit is

         d sin θ = m λ          m = ±1, ±2,…

a) Let's use the data to find the separation of the slits (d)

        d = m λ / sin θ

       d = 4 460 10⁻⁹ / sin 90

       d = 1.84 10⁻⁶ m

Now we can calculate the minimum wavelength for the third orcen (m = 3), the maximum angle must be 90 °

        λ  = d sin θ / m

        λ  = 1.84 10⁻⁶ sin 90 / 3

        λ  = 6.133 10⁻⁷ m

This is the last visible wavelength in this interference order, that is, we see the light with a shorter length than the calculated

b) To eliminate all the light of the fourth order (m = 4) let's use the minimum wavelength of visible range ( λ = 400 nm) from a 90 ° angle

          d = m  λ  / sin θ

          d = 4  400 10⁻⁹ / sin 90

         d = 1.60 10⁻⁶ m

If we reduce the slit to this value, the spectrum for the fourth order can be used.

c) It is change of separation is

          Δd = d₂ - d₁

          Δd = (1.84 - 1.60) 10⁻⁶ m

          Δd = 0.24 10⁻⁶ m