Question 34: Wave Optics

Question

In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is $$\lambda$$ is $$\mathrm{K}$$ units ($$\lambda$$ is the wavelength of light used). The percentage change in intensity at a point where the path difference is $$\frac{\lambda}{6}$$ and the above point is

Accepted Answer

In Young's double slit experiment for two waves of equal amplitude, the resultant intensity at a point is given by:
$$I = I_{	ext{max}} \cos^2\left(\frac{\phi}{2}ight),$$
where the phase difference is
$$\phi = \frac{2\pi \Delta}{\lambda}.$$
HereÃ¢ÂÂ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
$$I = I_{	ext{max}} 	imes 1 = I_{	ext{max}}.$$
It is given that this intensity is $$K$$ units, so
$$I_{	ext{max}} = K.$$
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
$$\cos^2\frac{\pi}{6} = \left(\frac{\sqrt{3}}{2}ight)^2 = \frac{3}{4}.$$
So the intensity in this case is
$$I = K 	imes \frac{3}{4} = 0.75K.$$
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
$$\left(\frac{0.25K}{K}ight) 	imes 100\% = 25\%.$$
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%.

Answer

75%

Answer

50%

Answer

4%

Answer

25%

Wave Optics

In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda is \mathrm{K} units (\lambda is the wavelength of light used). The percentage change in intensity at a point where the path difference is \frac{\lambda}{6}$ and the above point is

Choose the correct answer: