In a double-slit experiment, at a certain point on the screen the path difference between the two interfering waves is $$\frac{1}{8}$$th of a wavelength. The ratio of the intensity of light at that point to that at the center of a bright fringe is:

For a double-slit arrangement let the two slits give waves of equal individual intensity $$I_0$$. The general interference formula is first stated:

$$I \;=\; I_1 + I_2 + 2\sqrt{I_1I_2}\,\cos\phi,$$

where $$\phi$$ is the phase difference of the two waves at the observation point. Because both slits are identical, we have $$I_1 = I_2 = I_0$$, so the expression becomes

$$I \;=\; 2I_0 + 2I_0\cos\phi \;=\; 2I_0\bigl(1+\cos\phi\bigr).$$

Now we use the trigonometric identity $$1+\cos\phi = 2\cos^2\frac{\phi}{2}$$, giving

$$I \;=\; 2I_0 \times 2\cos^2\frac{\phi}{2} \;=\; 4I_0\cos^2\frac{\phi}{2}.$$

At the centre of a bright fringe (the central maximum) the two waves are in phase, so $$\phi = 0$$ and we obtain the maximum intensity

$$I_{\max} = 4I_0\cos^2 0 = 4I_0.$$

Our task is to find the intensity $$I$$ when the path difference $$\Delta$$ equals $$\dfrac{\lambda}{8}$$. Phase difference and path difference are connected by the relation

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

Substituting $$\Delta = \dfrac{\lambda}{8}$$ gives

$$\phi \;=\; \dfrac{2\pi}{\lambda}\,\Bigl(\dfrac{\lambda}{8}\Bigr) = \dfrac{2\pi}{8} = \dfrac{\pi}{4}.$$

Now we insert this value of $$\phi$$ into the intensity formula:

$$I \;=\; 4I_0\cos^2\frac{\phi}{2} = 4I_0\cos^2\!\Bigl(\dfrac{\pi}{4}\times\dfrac12\Bigr) = 4I_0\cos^2\!\Bigl(\dfrac{\pi}{8}\Bigr).$$

We are interested in the ratio of this intensity to the central maximum:

$$\dfrac{I}{I_{\max}} = \dfrac{4I_0\cos^2\!\bigl(\dfrac{\pi}{8}\bigr)}{4I_0} = \cos^2\!\Bigl(\dfrac{\pi}{8}\Bigr).$$

To evaluate $$\cos\!\bigl(\dfrac{\pi}{8}\bigr)$$ we employ the half-angle identity once more:

$$\cos\!\Bigl(\dfrac{\pi}{8}\Bigr) = \sqrt{\dfrac{1+\cos\!\bigl(\dfrac{\pi}{4}\bigr)}{2}} = \sqrt{\dfrac{1+\tfrac{1}{\sqrt2}}{2}} = \sqrt{\dfrac{1+\;0.7071}{2}} = \sqrt{\dfrac{1.7071}{2}} = \sqrt{0.8536} \approx 0.9239.$$

Squaring this value we obtain

$$\cos^2\!\Bigl(\dfrac{\pi}{8}\Bigr) \;=\; (0.9239)^2 \;\approx\; 0.8536.$$

Thus,

$$\dfrac{I}{I_{\max}} \approx 0.853.$$

Hence, the correct answer is Option A.