Question

Refer to Interactive Solution 27.9 for help in solving this problem. In a Young's double-slit experiment the separation distance y between the second-order bright fringe and the central bright fringe on a flat screen is 0.0151 m, when the light has a wavelength of 425 nm. Assume that the angles are small enough so that sin is approximately equal to tan . Find the separation y when the light has a wavelength of 560 nm.

Answer 1

Answer:

  y = 0.0199 m

Explanation:

The double slit experiment is described by the expression

          d sin θ = m λ

Where d is the separation of the slits, λ the wavelength and m an integer that gives the order of interference.

In this experiment the pattern is registered on a screen.

         tan θ = y / x

As the angles are very small

          tan θ = sin θ / cos θ = sin θ

We replace

         d y / x = m λ

Let's look with the initial data for the distance between the slits

          d / x = m λ / y

           m = 2

           d / x = 2 425 10⁻⁹ / 0.0151

           .d / x = 5.629 10⁻⁵ m

Now let's look for the spread if we change the wavelength

          λ = 560 nm = 560 10⁻⁹ m

          y = m λ   1 (d / x)

         

Let's calculate

          y = 2 560 10⁻⁹   1/ 5.629 10⁻⁵

          y = 1.9897 10⁻² m

          y = 0.0199 m