A double slit is illuminated with monochromatic light of wavelength 6.11 * 102 nm. The m = 0 and m = 1 bright fringes are separated by 1.8 cm on a screen which is located 3.5 m from the slits. What is the separation between the slits?

The position of a maximum given by interference created by a double slit is given by the formula:

d·sinα = m·λ

First, you need to find the angle α: you know the separation of the first two maxima and the distance of the slit from the screen, therefore, using trigonometry after transforming the units of measurements, you get:

y = 0.018 m

D = 3.5 m

α = tan⁻¹ (0.018/3.5) = 0.2945°

Now, solve for d in the first equation (λ = 6.11*10⁻⁷ m):

d = m·λ / sinα

= 1 · 6.11*10⁻⁷ / sin 0.2945

= 1.187 * 10⁻⁴ m

Therefore, the separation between the slits is 1.2 * 10⁻⁴ m.