Light of wavelength 680 nm falls on two slits and produces an interference pattern in which the fourth-order fringe is 47 mm from the central fringe on a screen 1.1 m away. What is the separation of the two slits?

To find the separation of the two slits, we can use the equation for the position of the fringes in a double-slit interference pattern:

y = (m * λ * D) / d

where:
- y is the distance between the central fringe and the mth-order fringe,
- λ is the wavelength of the light,
- D is the distance between the screen and the double-slit, and
- d is the separation of the two slits.

In this case, the fourth-order fringe is given as 47 mm and the wavelength of the light is 680 nm. The distance between the screen and the double-slit is 1.1 m.

Let's plug in these values into the equation and solve for d.

47 mm = (4 * 680 nm * 1.1 m) / d

First, let's convert the millimeters to meters:

47 mm = 0.047 m

Now, let's convert the wavelength from nanometers to meters:

680 nm = 680 * 10^(-9) m

Plugging in the values:

0.047 m = (4 * 680 * 10^(-9) m * 1.1 m) / d

Simplifying:

0.047 m * d = 4 * 680 * 10^(-9) m * 1.1 m

0.047 m * d = 2.9728 * 10^(-6) m

Finally, solving for d:

d = (2.9728 * 10^(-6) m) / 0.047 m

d ≈ 6.3277 * 10^(-5) m

Therefore, the separation of the two slits is approximately 6.3277 * 10^(-5) meters.