Question

In a Young's double slit experiment the slit is illuminated by a source having two wavelengths of 400 nm and 600 nm. If distance between slits, d=1 nm, and distance between the plane of the slit and screen, D=10 m the smallest distance from the central maximum where there is complete darkness is-

A. 2 mm
B. 3 mm
C. 12 mm
D. There is no such point

Answer 1

The question deals with finding a point of complete darkness in a Young's double slit experiment with two wavelengths which, due to the scales involved, does not have a solution for complete darkness.

Step-by-step explanation:

The subject question asks about the Young's double slit experiment involving two wavelengths of light. The student is looking to find the point of complete darkness, also known as a dark fringe, which occurs due to destructive interference. In the specific scenario described, with a slit separation (d) of 1 nm and wavelengths of 400 nm and 600 nm, no point of complete darkness would occur due to the extremely small separation compared to the wavelengths of light. This is because the condition for dark fringes (destructive interference) cannot be satisfied simultaneously for both wavelengths unless there is another factor, like a difference in the paths the light travels, which is not present in the given scenario.

The reference information shows examples of how to calculate various properties of double slit interference, such as the wavelength of light used in the experiment or the required slit separation to achieve a particular fringe separation. Each calculation requires knowledge of the equations governing interference patterns and the relationship between slit separation, wavelength, and the position of fringes on a screen.