Light of wavelength $600nm$passes though two slits separated by $0.20$$mm$and is observed on a screen $1.0m$behind the slits. The location of the central maximum is marked on the screen and labeled $y=0.$

a. At what distance, on either side of $y=0$, are the $m=1$bright fringes?

b. A very thin piece of glass is then placed in one slit. Because light travels slower in glass than in air, the wave passing through the glass is delayed by $5.0\times 10-16s$in comparison to the wave going through the other slit. What fraction of the period of the light wave is this delay?

c. With the glass in place, what is the phase difference $\mathrm{\Delta }{\varphi }_0$between the two waves as they leave the slits?$2$

d. The glass causes the interference fringe pattern on the screen to shift sideways. Which way does the central maximum move (toward or away from the slit with the glass) and by how far?

We use the formula to location of the initial maximum, and we discover that

$$\begin{gathered} y_1=\frac{1\lambda L}{d} \\ =\frac{6X{10}^{-7}X1m}{X\times {10}^{-4}} \\ =3X{10}^{-3} \\ =3\mathrm{mm} \end{gathered}$$

(b) Keep in mind that the time of a wave with a speed of $c$and a frequency $\lambda$of is $\tau =\frac{\lambda }{c}$. As a result, the ratio we're after is

$\frac{\mathrm{\Delta }t}{\tau }=\mathrm{\Delta }t\frac{c}{\lambda }$

In terms of numbers, we have

$$\begin{gathered} \frac{\mathrm{\Delta }t}{\tau }=\frac{5X{10}^{-16}sX3X{10}^{8}m/s}{6X{10}^{-7}m} \\ =\frac{1}{4} \end{gathered}$$

The aspect difference is simply our result amplified by $2\mathrm{\pi }.$

$$\begin{gathered} \mathrm{\Delta }\varphi =\frac{\mathrm{\Delta }t}{\tau }X2\pi \\ =2\pi \mathrm{\Delta }t\frac{c}{\lambda } \end{gathered}$$

In terms of numbers, we have

$\mathrm{\Delta }\varphi =\frac{\pi }{2}$

The axis will be shifted so that the time it takes for the light from the two slits to reach it is the same for each. Because one of the laser sources will arrive later, the new base must be placed near to the slit with the glass.

Consider the two right triangles generated by the light beam when there is no glass (the hypotenuses are denoted by $S$) and when the glass is added (the hypotenuses are denoted by $\left.S_1,S_2\right)$. It is clear that if we regard $S_1$to be the quickest distance,

$S_1=S-\mathrm{\Delta }S=S-c\mathrm{\Delta }t$

We may also infer from the Euclidean theorem that by examining this right triangle,

$\frac{d}{2}-x=\sqrt{S_1^2-L^2}$

As a conclusion, our unknown$x$will be

$x=\frac{d}{2}-\sqrt{S_1^2-L^2}$

Lastly, it should be self-evident.

$S=\sqrt{L^2+{\left(\frac{d}{2}\right)}^2}$

When we combine these, we obtain

$x=\frac{d}{2}-\sqrt{(S-\mathrm{\Delta }S{)}^2-L^2}$

Note that while we can give this statement in terms we know, it is much easier to calculate and then insert the values used in the aforementioned formula.