In the Young's double slit experiment the intensity produced by each one of the individual slits is . The distance between two slits is . The distance of screen from slits is . The wavelength of light is . The intensity of light on the screen in front of one of the slits is _____.
Correct answer:A
Standard Method
Given: Each slit alone produces intensity . Distance between slits , screen distance , and wavelength .
Find: The intensity at the point on the screen directly in front of one slit.
At the point directly in front of one slit, the point is not equidistant from both slits. Therefore, this point is neither a guaranteed central maximum nor a guaranteed minimum.
The nearer slit contributes intensity at that point. From the provided solution logic, the contribution of the other slit is taken as negligible at this location because of path difference and angular separation.
Therefore, the intensity on the screen in front of one slit is .
The correct option is A.
Assuming that every point on the screen in a double slit experiment gives either maximum or minimum intensity. This is wrong because strict maximum or minimum conditions require a specific path difference. Here the point is only directly in front of one slit, not equidistant from both slits.
Marking the answer as by adding slit intensities directly. This is wrong because in interference the resultant intensity depends on phase difference, not only on separate intensities. First identify the geometric position and interference condition.
Confusing the point in front of one slit with the central point of the screen. This is wrong because the central point is equidistant from both slits, whereas the point directly in front of one slit is laterally shifted from the center.
Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.
