Question

In Young's double slit experiment, 427 nm light gives a fourth-order bright fringe at a certain location on a flat screen. What is the longest wavelength of visible light that would produce a dark fringe at the same location

Answer 1

Answer:

λ' = 379.22nm

Explanation:

In order to find the longest wavelength that allows one to dark fringe coincides with the bright fringe, you use the following formulas:

$y_{bright}=\frac{m\lambda D}{d}$          (1)

$y_{dark}=(m+\frac{1}{2})\frac{\lambda' D}{d}$       (2)

y-dark and y-bright are the positions of dark fringes and bright fringes respectively.

m: order of the fringe (dark or bright) = 4

D: distance to the screen

d: distance between slits

λ: light for y-bright= 427nm

λ': second light for y-dark = ?

By the information of the statement you know that y-bright = y-dark.

You divide equation (2) into the equation (1) and solve for λ':

$\frac{y_{dark}}{y_{bright}}=1=\frac{(m+1/2)\lambda'}{m\lambda}\\\\\lambda'=\frac{m\lambda}{m+1/2}$  (3)

Finally, you solve the equation (3) by replacing the values of m and the wavelength:

$\lambda'=\frac{4(427nm)}{4+1/2}=379.55nm$

The longest wavelength which produces the fourth dark fringe in the same location for the fourth bright fringe of the first wavelength is 379.22nm