Question

The diameter of a thin wire is measured in a physics laboratory by a student. The wire is held vertically in a holding frame in front of the laser beam. The laser light diffracts on the wire and produces a diffraction pattern on a white screen. The diffraction pattern is shown in the figure. The pattern is centered around the origin.

The diameter of a thin wire is measured in a physics laboratory by a student The wire is held vertically in a holding frame in front of the laser beam. The laser light diffracts on the wiré and produces a diffraction pattern on a white screen The diffraction pattern is shown in the figure. The pattern is centered around the origin. 4 2 0 -1 -2 -3 8-6 5 -43 -2 1 02 3 45 6 7 8 x (cm) The wavelength of the laser light is 455 nm, and the screen is 1.47 m away from the wire. What is the diameter of the wire? (Hint: one of the higher order diffraction minima lines up with a well defined x-value. Also, it is perfectly safe to use the s mall angle approximation: sin(θ) = tan(θ)). Submit Answer Incorrect. Tries 4/12 Previous Tries

Answer 1

We know that in the diffraction experiment for dark band
dSin$\theta$ = n$\lambda$
where n is the order and $\lambda$ is the wavelength of the light.
for small angle we can write Sin$\theta$ = tan $\theta$
tan$\theta$ = y/D
where y is the distance of the dark band from the central band and D is the distance of the screen
d*(y/D) = n$\lambda$
d = n$\lambda$ D/y
where d is the slit of width . In this case the diameter of wire must be equal to slit width.
From the figure we can find third dark band is at 4 cm
hence n = 3 , y = 4 cm
d = 3*(455*10^-9)*1.47 / (4*10^-2)
d = 5.01*10^-5 m