t),$$

where the phase difference is

$$\phi = \frac{2\pi \Delta}{\lambda}.$$

Here’s how to solve the problem step by step:

For the point where the path difference is $$\Delta = \lambda$$:

The phase difference is

$$\phi = \frac{2\pi \lambda}{\lambda} = 2\pi.$$

Therefore,

$$\frac{\phi}{2} = \pi$$

and since

$$\cos \pi = -1,$$

we have

$$\cos^2\pi = 1.$$

Thus, the intensity is

$$I = I_{\text{max}} \times 1 = I_{\text{max}}.$$

It is given that this intensity is $$K$$ units, so

$$I_{\text{max}} = K.$$

Now, for the point where the path difference is $$\Delta = \frac{\lambda}{6}$$:

The phase difference becomes

$$\phi = \frac{2\pi (\lambda/6)}{\lambda} = \frac{2\pi}{6} = \frac{\pi}{3}.$$

Then,

$$\frac{\phi}{2} = \frac{\pi}{6}.$$

The cosine of $$\frac{\pi}{6}$$ is

$$\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2},$$

which means

$$\cos^2\frac{\pi}{6} = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}.$$

So the intensity in this case is

$$I = K \times \frac{3}{4} = 0.75K.$$

Comparing the two intensities:

The intensity decreases from $$K$$ to $$0.75K$$.

The change in intensity is

$$\Delta I = K - 0.75K = 0.25K,$$

which, as a percentage of the original intensity $$K$$, is

$$\left(\frac{0.25K}{K}\right) \times 100\% = 25\%.$$

Thus, the percentage change in intensity at a point where the path difference is $$\frac{\lambda}{6}$$ compared to the point where the path difference is $$\lambda$$ is 25%.

The correct answer is Option D: 25%.

Wave Optics

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