In a Young's double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is $$\dfrac{1}{8}$$ th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to:

In a Young’s double-slit experiment two coherent waves originating from the two slits reach a point on the screen with a certain path difference. Let this path difference be denoted by $$\Delta x$$. For the present point we are told that

$$\Delta x=\dfrac{\lambda}{8},$$

where $$\lambda$$ is the wavelength of the monochromatic light being used.

Whenever two waves of the same amplitude interfere, the intensity at any point on the screen is obtained from the interference formula

$$I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi.$$

Because both slits are identical, the individual intensities are the same, i.e. $$I_1 = I_2 = I_s$$. Substituting $$I_1 = I_2 = I_s$$ in the above expression we get

$$I = I_s + I_s + 2\sqrt{I_s I_s}\cos\phi$$

$$\;\; = 2I_s + 2I_s\cos\phi$$

$$\;\; = 2I_s(1+\cos\phi).$$

The phase difference $$\phi$$ between the two waves is related to the path difference $$\Delta x$$ through the relation

$$\phi = \dfrac{2\pi}{\lambda}\,\Delta x.$$

Substituting $$\Delta x = \dfrac{\lambda}{8}$$, we obtain

$$\phi = \dfrac{2\pi}{\lambda}\left(\dfrac{\lambda}{8}\right) = \dfrac{2\pi}{8} = \dfrac{\pi}{4}.$$

Now, the cosine of this phase difference is

$$\cos\phi = \cos\left(\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2} \approx 0.707.$$

Putting this value back into the intensity expression, the intensity at the given point becomes

$$I = 2I_s\left(1 + \dfrac{\sqrt{2}}{2}\right).$$

The intensity at the centre of the central bright fringe, commonly called the maximum intensity $$I_{\text{max}}$$, occurs when the two waves meet in perfect phase (that is, $$\phi = 0$$). Then the formula gives

$$I_{\text{max}} = 2I_s(1 + \cos 0) = 2I_s(1 + 1) = 4I_s.$$

We now require the ratio of the intensity at the given point to the maximum intensity. Taking the ratio, we have

$$\dfrac{I}{I_{\text{max}}} = \dfrac{2I_s\left(1 + \dfrac{\sqrt{2}}{2}\right)}{4I_s} = \dfrac{1 + \dfrac{\sqrt{2}}{2}}{2}.$$

Simplifying the numerator first, write $$1 = \dfrac{2}{2}$$, so

$$1 + \dfrac{\sqrt{2}}{2} = \dfrac{2 + \sqrt{2}}{2}.$$

Substituting this back into the ratio,

$$\dfrac{I}{I_{\text{max}}} = \dfrac{\dfrac{2 + \sqrt{2}}{2}}{2} = \dfrac{2 + \sqrt{2}}{4}.$$

Because $$\sqrt{2} \approx 1.414$$, the numerator becomes

$$2 + 1.414 = 3.414.$$

Dividing by 4 gives

$$\dfrac{3.414}{4} \approx 0.8535.$$

This numerical value is closest to 0.85 among the choices provided.

Hence, the correct answer is Option B.