In a Young's double slit experiment, light of 500 nm is used to produce and interference pattern. When the distance between the slits is 0.05 mm, the angular width (in degree) of the fringes formed on the distance screen is close to:

We have a Young’s double-slit arrangement. For any two consecutive bright (or dark) fringes, the angular separation is given by the standard result of interference:

$$\Delta\theta \;=\;\frac{\lambda}{d}$$

where $$\lambda$$ is the wavelength of light and $$d$$ is the centre-to-centre separation of the two slits.

The wavelength given is 500 nm. First we convert it completely into SI units:

$$500\ \text{nm}=500\times10^{-9}\ \text{m}=5\times10^{-7}\ \text{m}$$

The slit separation is 0.05 mm, which we again express in metres:

$$0.05\ \text{mm}=0.05\times10^{-3}\ \text{m}=5\times10^{-5}\ \text{m}$$

Now we substitute these numerical values into the formula for $$\Delta\theta$$:

$$\Delta\theta \;=\;\frac{\lambda}{d}\;=\;\frac{5\times10^{-7}\ \text{m}}{5\times10^{-5}\ \text{m}}$$

The powers of ten and the numerical factors simplify very neatly:

$$\Delta\theta \;=\;\frac{5}{5}\times10^{(-7)-(-5)}\;=\;1\times10^{-2}\ \text{radian}$$

That is

$$\Delta\theta =0.01\ \text{radian}$$

To make comparison with the options easier, we convert this radian value to degrees, using the relation $$1\ \text{radian}=57.3^\circ$$ (approximately):

$$\Delta\theta =0.01\times57.3^\circ =0.573^\circ$$

On rounding to two significant figures, this becomes $$0.57^\circ$$, which matches the option provided.

Hence, the correct answer is Option B.