Answers to Optics Practical 3

Question 1 §

Using an interference equation and substituting the values in the question converted into metres: $y_{\text{bright}}=L\frac{m\lambda }{d}⟹0.82\text{m}=12\cdot \frac{1\text{m}\cdot (633\cdot 10^{-9})\text{m}}{d}$

$$⟹d=\frac{12\cdot 633\cdot 10^{-9}}{0.82}\text{m}=\frac{1899}{205000000}\text{m}\approx 9.26\cdot 10^{-6}\text{m}$$

Thus the separation of slits is approximately $9.26\cdot 10^{-6}\text{m}$.

In principle, you would observe $n$ interference maxima, where $n$ occurs when the angle from the slit to the maxima is $90^{\circ }$, so that $n$ can be solved for in the following equation:

$$(9.26\cdot 10^{-6}\text{m})\cdot \sin 90^{\circ }=n\cdot (633\cdot 10^{-9})\text{m}$$

$$⟹n=\frac{9.26\cdot 10^{-6}\text{m}}{633\cdot 10^{-9}\text{m}}=\frac{9260}{633}\approx 14.63\therefore n=14$$

This is just in one direction, so this must be doubled and the central maxima must be added, so $n=29$. However, in practice this may not be observed as each maxima gets gradually dimmer.

Question 2 §

Using an interference equation and substituting the values in the question converted into metres:

$$y_{\text{bright}}=L\frac{m\lambda }{d}⟹y_{\text{bright}}=1\cdot \frac{1\cdot (500\cdot 10^{-9})}{1\cdot 10^{-2}}\text{m}=500\cdot 10^{-7}\text{m}=5\cdot 10^{-5}\text{m}$$

Thus, the spacing between each maxima would be $5\cdot 10^{-5}\text{m}$.

Question 3 §

Light is shining normally to the $n_1-n_2$ interface, thus the angle is $90^{\circ }$. Using the equation given in the question,

$$I={\left(\frac{n_1-n_2}{n_1+n_2}\right)}^2\cdot I_0$$

we can calculate the intensity between an air-glass interface in terms of the incidence intensity, using $n_1=1$ and $n_2=1.5$:

$$I_1={\left(\frac{1-1.5}{1+1.5}\right)}^2\cdot I_0=0.04\cdot I_0$$

We can then continue this equation to find the following intensities whilst the light travels through the slab of glass:

$$I_2={\left(\frac{1.5-1}{1.5+1}\right)}^2\cdot (0.96\cdot I_0)=0.0384\cdot I_0$$

Thus, the remaining intensity out of the slab of glass would be $0.96-0.0384$, therefore:

$$I_{\text{final}}=0.9216\cdot I_{\text{initial}}$$