A transparent slab is inserted in front of one slit of a Young's double-slit arrangement and shifts the entire interference pattern by 3.5 fringe widths. What intensity condition is produced at the point that was originally the central maximum?
Options

Question

A transparent slab is inserted in front of one slit of a Young's double-slit arrangement and shifts the entire interference pattern by 3.5 fringe widths. What intensity condition is produced at the point that was originally the central maximum?

Options

G1 Physics · Accepted Answer

A shift of one fringe width corresponds to one wavelength of added path. Thus a shift of 3.5β corresponds to 3.5λ = 7λ ÷ 2. This is a destructive-interference condition. Therefore the original central point becomes a minimum. Correct option: B.