Question

Light of a single color hits the wall after traveling 1.40 m from a small sliver of your window of width 0.600 mm. A diffraction pattern appears on your wall with a central maximum width of 1.15 mm. What is the wavelength of the light

Answer 1

Condition for diffraction maximum is

$dsin\theta = m\lambda$

Here,

d = Distance between slits

m =Order of interference, or any integer which represent the number of repetition of the spectrum

$\lambda$ = Wavelength

For small angles we have that

$sin\theta = tan\theta = \frac{y}{L}$

L = Distance of the Screen

y = Position on the screen

At the same time we have that the distance of the edge of central maximum is

$w = 2y$

$y = \frac{w}{2L}$

Replacing all in the first equation we have

$d(\frac{w}{2L}) = m\lambda$

Remember that for the maximum value to be given, then the order of interference must be 1, replacing with the other values we will have to,

$(0.600*10^{-3}) (\frac{1.15*10^{-3}}{2(1.4)}) = (1) \lambda$

$\lambda =2.464*10^{-7}m$

$\lambda = 246.4nm$

Therefore the wavelength of the light is 246.4nm