Question

g We have studied diffraction from a single slit, where light is sent through a thin opening. A similar phenomena occurs when light bends around a thin object, like a human hair. Here the width of the hair plays the role of the width of the single slit. Measurements found that when a beam of light of wavelength 632.8 nm was shone on a single strand of hair, the first dark fringe on either side of the central bright spot were 5.22 cm apart. If the screen is 1.25 meters away, how thick was this strand of hair?

Answer 1

Answer:

$3.031\times 10^{-5}\ \text{m}$

Explanation:

$y$ = Distance between central maxima and first minimum

m = Order = 1

d = Thickness of hair

$\lambda$ = Wavelength = 632.8 nm

L = Distance between light source and screen = 1.25 m

Width of central maximum is given by

$2y=5.22\times 10^{-2}\\\Rightarrow y=\dfrac{5.22\times 10^{-2}}{2}\\\Rightarrow y=0.0261\ \text{m}$

Distance between central maxima and first minimum is given by

$y=L\tan\theta_{min}\\\Rightarrow \tan\theta_{min}=\dfrac{y}{L}\\\Rightarrow \tan\theta_{min}=\dfrac{0.0261}{1.25}\\\Rightarrow \theta_{min}=\tan^{-1}0.02088\\\Rightarrow \theta_{min}=1.1962^{\circ}$

Since $\theta$ is small $\tan\theta_{min}=\sin\theta_{min}$

$\sin\theta_{min}=\dfrac{m\lambda}{d}\\\Rightarrow d=\dfrac{m\lambda}{\sin\theta}\\\Rightarrow d=\dfrac{1\times 632.8\times 10^{-9}}{\sin1.1962^{\circ}}\\\Rightarrow d=3.031\times 10^{-5}\ \text{m}$

The strand of hair is $3.031\times 10^{-5}\ \text{m}$ thick.