Question

Consider red light with a wavelength of 692 nm in air that is incident onto two slits such that it gives diffraction fringes that are 1.00 mm apart on a screen at a distance L from the slits. If the screen is moved back by an additional 4.00 cm, the fringes become 1.20 mm apart. What is the separation between the slits?

Answer 1

Answer:

The separation between slits is:

$d=0.1384 \, mm$

Explanation:

The distance between adjacent maxima is given by the following expression:

$\Delta y=\frac{\lambda L}{d}$

For the distance between maxima being 1mm (let's solve all this problem using milimeters) we have the following:

$1=\frac{L\lambda}{d}$ (1)

For the second condition in which the screen is moved 40 mm away we have the following:

$1.20=\frac{(L+40)\lambda}{d}$ (2)

Using equation 1 we can write:

$L=\frac{L}{d}$

Plugging the value of L in equation 2 brings us to

$1.20=\frac{\lambda\left(\frac{d}{\lamda}+40\right)}{d}$

$\implies 1.20=\frac{d+40\lambda}{d}$

$\implies 1.20d=d+40\lambda$

$\implies d=\frac{40\lambda}{0.20}=0.1384\, mm$

We could also as a bonus find L which is just

$L=\frac{d}{\lambda}$

I'll leave that up to you! Good luck!