Adouble 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 / sin0.2945
   = 1.187×10⁻⁴ m

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

    5.5 n======    assuming that      we have      m = 0.145 kg    a = 38 m/s²therefore    5.51 n, but two significant figures.final answer: 5.5 n