In a Young's double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern is observed on a screen kept 1.5 m away. The separation between the successive bright fringes on the screen is:

We know that, in Young’s double slit experiment, the linear separation between two successive bright (or dark) fringes on the screen is called the fringe width $$\beta$$ and is given by the formula

$$\beta \;=\;\dfrac{\lambda\,D}{d}$$

where

$$\lambda$$ = wavelength of the monochromatic light,

$$D$$ = distance between the double-slit plane and the screen,

$$d$$ = separation between the two slits.

First, we convert every quantity to SI units:

Wavelength: $$\lambda = 589 \text{ nm} = 589 \times 10^{-9}\ \text{m}$$

Slit separation: $$d = 0.15 \text{ mm} = 0.15 \times 10^{-3}\ \text{m}$$

Screen distance: $$D = 1.5\ \text{m}$$ (already in SI units).

Now we substitute these values into the formula:

$$ \beta = \dfrac{\lambda\,D}{d} = \dfrac{(589 \times 10^{-9}\ \text{m}) \;(1.5\ \text{m})}{0.15 \times 10^{-3}\ \text{m}} $$

First we multiply the numerical values in the numerator:

$$589 \times 1.5 = 883.5$$

So,

$$ \beta = \dfrac{883.5 \times 10^{-9}\ \text{m}^{2}}{0.15 \times 10^{-3}\ \text{m}} $$

Next, we divide the powers of ten separately. We have

$$\dfrac{10^{-9}}{10^{-3}} = 10^{-9 - (-3)} = 10^{-6}$$

and for the numerical division:

$$\dfrac{883.5}{0.15} = \dfrac{883.5 \times 100}{15} = \dfrac{88350}{15} = 5890$$

Hence,

$$ \beta = 5890 \times 10^{-6}\ \text{m} = 5.890 \times 10^{-3}\ \text{m} $$

To convert metres to millimetres, we recall that $$1\ \text{m} = 1000\ \text{mm}$$, so

$$ \beta = 5.890 \times 10^{-3}\ \text{m} \times 1000\ \dfrac{\text{mm}}{\text{m}} = 5.890\ \text{mm} \approx 5.9\ \text{mm}. $$

Hence, the correct answer is Option C.