In Young\'s double-slit experiment, 632.8 nm light from a HeNe laser passes thro

Question

In Young's double-slit experiment, 632.8 nm light from a HeNe laser passes through the two slits and is projected on a screen. As expected, a central maximum (constructive interference) is observed at the center point on the screen. Now, a very thin piece of plastic with an index of refraction n=1.48 covers one of the the slits such that the center point on the screen, instead of being a maximum, is dark.

Part A

Determine the minimum thickness of the plastic.

Hint: Recall that the index of refraction changes the speed of light as v=c/n, thereby decreasing the wavelength of light in the plastic.

Explanation / Answer

To change the center point from constructive interference to destructive interference, the phase shift

produced by the introduction of the plastic must be equivalent to half a wavelength. The wavelength of the light

is shorter in the plastic than in the air, so the number of wavelengths in the plastic must be 1/2 greater than the

number in the same thickness of air. The number of wavelengths in the distance equal to the thickness of the plate

is the thickness of the plate divided by the appropriate wavelength.

t/ lamda_n - t/ lamda = 1/2

t/ lamda/n - t/ lamda = 1/2

nt/ lamda - t/lamda = 1/2

t /lamda * ( n-1) = 1/2

t = lamda/ 2( n-1)

= 632.8 nm/ 2( 1.48-1)

=659.16 nm

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