Question

In Fig. 35-45, two microscope slides touch at one end and are separated at the other end. When light of wavelength 500 nm shines vertically down on the slides, an overhead observer sees an interference pattern on the slides with the dark fringes separated by 1.2 mm. What is the angle between the slides?

Short answer

The angel between the slides is $0.012°$.

Step-by-step solution

1.Introduction

Microscope is an instrument that can be used to observe small objects, even cells. The image of an object is magnified through at least one lens in the microscope. This lens bends light toward the eye and makes an object appear larger than it actually is.

Angle between the slides

The vertical change between the centre of one dark band and the next can be calculated by using the following relation

$\Delta y=\frac{\lambda }{2}$

Here, $\lambda$ is the wavelength of the light

The angle between the slides can be calculated by using the following relation

$\tan \theta =\frac{\Delta y}{\Delta x}$

Here, $\Delta x$is the horizontal separation between the dark fringes.

Vertical change between the fringes

The vertical change between the centre of one dark band and the next dark band is,

$\Delta y=\frac{\lambda }{2}$

Substitute 500 mm for $\lambda$.

$\begin{array}{rcl}\Delta y& =& \frac{\left(500\text{mm}\right)\left(\frac{1\times {10}^{-9}\text{m}}{1\text{mm}}\right)}{2}\\ & =& \frac{500\times {10}^{-9}\text{m}}{2}\\ & =& 2.50\times {10}^{-7}\text{m}\end{array}$

Determination of the angle between the slides

The angel between the slides is,

$\theta =\frac{\Delta y}{\Delta x}$

Substitute $2.50\times {10}^{-7}\text{m}$for $\Delta y$and $1.2\mathrm{mm}$for $\Delta \text{x}$.

$\begin{array}{rcl}\theta & =& \frac{2.50\times {10}^{-7}\text{m}}{\left(1.2\text{mm}\right)\left(\frac{1.0\times {10}^{-3}\text{m}}{1.0\text{mm}}\right)}\\ & =& \left(2.08\times {10}^{-4}\text{rad}\right)\left(\frac{57.2957795°}{1\text{rad}}\right)\\ & =& 0.012°\end{array}$

Therefore, the angel between the slides is $0.012°$.