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.