Chapter 37, Problem 37.60AP

Consider the double-slit arrangement shown in Figure P37.60, where the slit separation is d and the distance from the slit to the screen is L. A sheet of transparent plastic having an index of refraction n and thickness t is placed over the upper slit. As a result, the central maximum of the interference pattern moves upward a distance y′ Find y′.

To determine

The position $y^{\prime }$ of the central maxima of the interference pattern.

Answer to Problem 37.60AP

The position $y^{\prime }$ of the central maxima of the interference pattern is $\frac{\left(n-1\right)Lt}{\sqrt{\left(d^2-{\left(n-1\right)}^2{t}^2\right)}}$.

Explanation of Solution

Given info: The separation between the slits is $d$, the distance from the slit to the screen is $L$, the index of refraction of transparent plastic sheet is $n$, the thickness of the plastic sheet is $t$.

The path difference due to the slit is given by,

${\delta }_{\text{slit}}=d\sin \theta$ (1)

Here,

${\delta }_{\text{slit}}$ is the difference due the difference in path of two waves due to slits

$d$ is the separation between the slits.

$\theta$ is the angle subtended by the point on the screen.

The value of sine of angle is,

$\sin \theta =\frac{y^{\prime }}{\sqrt{L^2+{\left(y^{\prime }\right)}^2}}$

Here,

$y^{\prime }$ is the position the new central maxima on the screen.

$L$ is the distance of the screen from the slits.

Substitute $\frac{y^{\prime }}{\sqrt{L^2+{\left(y^{\prime }\right)}^2}}$ for $\sin \theta$ in equation (1).

${\delta }_{\text{slit}}=d\left(\frac{y^{\prime }}{\sqrt{L^2+{\left(y^{\prime }\right)}^2}}\right)$

The path difference due to the transparent plastic sheet is,

${\delta }_{\text{slit}}=\left(n-1\right)t$

Here,

${\delta }_{\text{slit}}$ is the optical path difference due the plastic sheet.

$n$ is the refractive index of the plastic sheet.

$t$ is the thickness of the plastic sheet.

The net path difference is given by,

${\delta }_{\text{net}}={\delta }_{\text{slit}}-{\delta }_{\text{pt}}$

Substitute $d\left(\frac{y^{\prime }}{\sqrt{L^2+{\left(y^{\prime }\right)}^2}}\right)$ for ${\delta }_{\text{slit}}$ and $\left(n-1\right)t$ for ${\delta }_{\text{pt}}$ in the above equation.

${\delta }_{\text{net}}=d\left(\frac{y^{\prime }}{\sqrt{L^2+{\left(y^{\prime }\right)}^2}}\right)-\left(n-1\right)t$

The condition of central maxima is,

${\delta }_{\text{net}}=0$

Substitute $d\left(\frac{y^{\prime }}{\sqrt{L^2+{\left(y^{\prime }\right)}^2}}\right)-\left(n-1\right)t$ for ${\delta }_{\text{net}}$ in the above equation.

$\begin{array}{c}d\left(\frac{y^{\prime }}{\sqrt{L^2+{\left(y^{\prime }\right)}^2}}\right)-\left(n-1\right)t=0\\ d\left(\frac{y^{\prime }}{\sqrt{L^2+{\left(y^{\prime }\right)}^2}}\right)=\left(n-1\right)t\\ d{y}^{\prime }=\left(\sqrt{L^2+{\left(y^{\prime }\right)}^2}\right)\left(n-1\right)t\end{array}$

Take square on both sides of the above equation.

$\begin{array}{c}{d}^2{\left(y^{\prime }\right)}^2=\left(L^2+{\left(y^{\prime }\right)}^2\right){\left(\left(n-1\right)t\right)}^2\\ {\left(y^{\prime }\right)}^2\left(d^2-{\left(\left(n-1\right)t\right)}^2\right)=L^2{\left(\left(n-1\right)t\right)}^2\\ {\left(y^{\prime }\right)}^2=\frac{L^2{\left(\left(n-1\right)t\right)}^2}{\left(d^2-{\left(\left(n-1\right)t\right)}^2\right)}\end{array}$

Simplify the above equation for $y^{\prime }$.

$\begin{array}{c}{y}^{\prime }=\sqrt{\frac{L^2{\left(\left(n-1\right)t\right)}^2}{\left(d^2-{\left(\left(n-1\right)t\right)}^2\right)}}\\ =\frac{L\left(\left(n-1\right)t\right)}{\sqrt{\left(d^2-{\left(\left(n-1\right)t\right)}^2\right)}}\\ =\frac{\left(n-1\right)Lt}{\sqrt{\left(d^2-{\left(n-1\right)}^2{t}^2\right)}}\end{array}$

Conclusion:

Therefore, the position $y^{\prime }$ of the central maxima of the interference pattern is $\frac{\left(n-1\right)Lt}{\sqrt{\left(d^2-{\left(n-1\right)}^2{t}^2\right)}}$.