Chapter 24, Problem 50P

To determine

The thickness of the foil.

Answer to Problem 50P

  $80.1\mu \text{m}$ .

Explanation of Solution

Given:

Wavelength, $\lambda =589nm$

Number of fringes shift, $N=272$

Formula used:

It is known that: $\Delta L=\frac{\text{N}\times \lambda }{2}$

Where,

  $\Delta L$ is thickness.

  $\text{λ}$ is wavelength of the light.

  $\text{N}$ is the number of fringes shift.

Calculation:

Substitute the data into the formula $\Delta L\text{=}\frac{\text{N×λ}}{\text{2}}$ to get the thickness of the foil,

  $\begin{array}{c}\Delta L\text{=}\frac{\text{N×λ}}{\text{2}}\\ \text{=}\frac{\text{272×589}\text{nm}}{\text{2}}\\ \text{=80}\text{.1}\text{μm}\end{array}$

Conclusion:

The thickness of the foil is $\text{80}\text{.1}\text{μm}$ .