Q 12-10-089JEE MainJEE Main 2023 (8 Apr, Shift 2)Easy
The width of fringe is $2$ mm on the screen in a double slit experiment for the light of wavelength of $400$ nm. The width of the fringe for the light of wavelength $600$ nm will be
Answer: (D) $3\ \text{mm}$
$\beta\propto\lambda$: $\beta=2\times\dfrac{600}{400}=3$ mm.
Solution by Sreeraj P, M.Sc Physics