Question

A very thin sheet of plastic (n=1.60) covers one slit of a double-slit apparatus illuminated by 640nm light. The center point on the screen, instead of being a maximum, is dark.

What is the minimum thickness of the plastic?
t= _______ nm

Answer 1

Refractive index of the plastic sheet, \(n=1.60\) The wavelength in the plastic, \(\lambda / n=\frac{640}{1.6}\) \(=400 \mathrm{~nm}\)

The distance thru the plastic must be \(1 / 2\) wavelength longer than thru the same thickness of air

So \(m \lambda=\left(m+\frac{1}{2}\right) \frac{\lambda}{n}\)

where \(m\) is the number of wavelengths to sati sfy this condition \(m=\left(m+\frac{1}{2}\right) \frac{1}{n}\)

\(n \times m-m=\frac{1}{2}\)

\(m \times(1.6-1)=0.5\)

\(\Rightarrow m=0.8333\)

So the thickness of the plastic, \(t=0.8333 \times(640 \mathrm{~nm})\)

$$ =533 \mathrm{~nm} $$

As a check if the same thickness of air contains \(5 / 6 \lambda\) then

the plastic must contain an extra \(\frac{1}{2} \times\) (Wavelength) \(\Rightarrow \frac{8}{6} \lambda=\frac{4}{3} \lambda\), where \(\lambda\) of the plastic \(=400 \mathrm{~nm}\)

$$ =533 \mathrm{~nm} $$