The correct answer is: B. I04\frac{I_0}{4}
In Young's double slit experiment, the intensity of light on the screen is given by the interference pattern formed by the light from the two slits. The intensity II at any point on the screen is determined by the interference equation:
I=I0cos2(πdsinθλD)I = I_0 \cos^2 \left( \frac{\pi d \sin \theta}{\lambda D} \right)
Where:
• I0I_0 is the maximum intensity at the central maximum.
• dd is the distance between the slits.
• λ\lambda is the wavelength of the light.
• DD is the distance from the slits to the screen.
• θ\theta is the angle relative to the central axis.
Step 1: Determine the position of the point in front of one of the slits:
• We are asked to find the intensity in front of one of the slits. This corresponds to a point where the angle θ=0\theta = 0 (directly in front of one of the slits).
When θ=0\theta = 0, the condition for constructive or destructive interference is no longer in play, and we can directly use the intensity formula.
Step 2: Intensity for a point in front of one of the slits:
• For the case where θ=0\theta = 0, the path difference between the two slits is zero, and the light from both slits will interfere constructively (in phase).
• The interference factor will be maximized for this point, but because the two slits are producing coherent light sources, the resultant intensity is not I0I_0, but instead it is due to the superposition of the two sources.
Since the slits are in phase and the distance from the screen is given as D=10dD = 10d, this results in a scenario where the interference between the two slits contributes to the intensity being halved due to the path difference being effectively zero.
Thus, the intensity at this point is:
I=I04I = \frac{I_0}{4}
This accounts for the diffraction and interference patterns formed by the slits.
Conclusion:
The intensity of light at the point in front of one of the slits will be I04\frac{I_0}{4}.
Thus, the correct answer is B. I04\frac{I_0}{4}.
