Question

In Fig. 35-45, a broad beam of light of wavelength 620 nm is sent directly downward through the top plate of a pair of glass plates touching at the left end. The air between the plates acts as a thin film, and an interference pattern can be seen from above the plates. Initially, a dark fringe lies at the left end, a bright fringe lies at the right end, and nine dark fringes lie between those two end fringes. The plates are then very gradually squeezed together at a constant rate to decrease the angle between them. As a result, the fringe at the right side changes between being bright to being dark every 15.0 s.

(a) At what rate is the spacing between the plates at the right end being changed?

(b) By how much has the spacing there changed when both left and right ends have a dark fringe and there are five dark fringes between them?

Short answer

(a) The rate of change of this spacing is $10.33\text{nm/s}$.

(b) The change in the spacing is $1085nm$.

Step-by-step solution

Introduction

Wavelength is defined as the distance between successive crests of a wave, especially points in a sound wave or electromagnetic wave.

The rate at which the spacing between the plates at the right end being changed

(a)

Here initially there are 9 dark fringes between two ends.

Therefore we should have $L_R=\left(m+\frac{1}{2}\right)\frac{\lambda }{2n}$

Here $n=1,m=9$ (since 9th dark fringe)

Wave length $\lambda =620\text{nm}$

$$\begin{gathered} L_R=\left(9+\frac{1}{2}\right)\left(\frac{620\text{nm}}{2\left(1\right)}\right) \\ =2945 \end{gathered}$$

But in 15 s every bright fringe is changing to dark fringe.

So after 15 s the 9th dark fringe is changing to 9th bright fringe.

Therefore for bright fringe,

$$\begin{gathered} L_R^{\text{'}}=m\frac{\lambda }{2n} \\ =\frac{\left(9\right)\left(620\text{nm}\right)}{2\left(1\right)} \\ =2790 \end{gathered}$$

Therefore changing in the width of the left edge is $\Delta L_R=L_R-L_R^{\text{'}}$

$$\begin{gathered} =2745\text{nm - 2790}\text{nm} \\ \text{= 155}\text{nm} \end{gathered}$$

This change takes place in $\Delta t=15\text{s}$

The rate of change of this spacing is

$$\begin{gathered} \frac{\Delta L_{\tau }}{\Delta t}\frac{155\text{nm}}{15\text{s}} \\ =10.33\text{nm/s} \end{gathered}$$

Hence, the rate of change of this spacing is $10.33nm/s$.

Step 3: The change in spacing when both left and right ends have a dark fringe

(b)

Here there are 5 dark fringes between left and right ends and each end have dark fringes. Therefore there are 7 dark fringes.

Therefore $m=6$

role="math" localid="1663148654843" $$\begin{gathered} L_R^{\text{'}}=\left(6\right)\frac{\lambda }{2n} \\ =\frac{\left(620\text{nm}\right)}{\left(2\right)\left(1\right)} \\ =1860\text{nm} \end{gathered}$$

Change in the spacing is

$$\begin{gathered} L_R-L_R^{\text{'}}=2945\text{nm - 1860}\text{nm} \\ \text{= 1085}\text{nm} \end{gathered}$$

Hence, the change in the spacing is $1085nm$.