A $5.0-\mathrm{cm}$-thick layer of oil is sandwiched between a $1.0-\mathrm{cm}$ thick sheet of glass and a $2.0-\mathrm{cm}$-thick sheet of polystyrene plastic. How long (in ns) does it take light incident perpendicular to the glass to pass through this $8.0-\mathrm{cm}$-thick sandwich?

1. The medium's degree for refraction $n$was described as the proportion

$n=\frac{c}{v}$ $\left(1\right)$

The maximum speed into vacuum was $c=3.0\times {10}^8\mathrm{m}/\mathrm{s}$and also the light speed inside the media is $v$.

2. The Ray Structure of Light: Light passes at $v=c/n$via neat lines called light ray. The light ray will persist indefinitely except if it interacts with substance, in which case it will reflects, refract, disperse, or be consumed. Light rays are emitted by things. Rays are sent out in every dimensions from every spot on the item. When divergent rays are gathered either by pupil and centered just on retina, the person perceives an item (or a picture). If lens, reflectors, and apertures are bigger over $\approx 1\mathrm{mm}$, ray optics is acceptable.

3. The molecule's maximum pace was equivalent here to proportion of the number path that moves divided by the sum period that goes a certain length:

$v_{\mathrm{avg}}=\frac{d}{\mathrm{\Delta }t}$ $\left(2\right)$

$n_{\text{oil}}=1.46$would be the reflectance for oils.

$n_{\text{glass}}=1.50$would be the reflectance for glasses.

$n_{\text{plastic}}=1.59$would be the reflectance for polystyrene plastics.

$d_{\mathrm{oil}}=5.0\mathrm{cm}$would be the density of a oil layer.

$d_{\text{glass}}=1.0\mathrm{cm}$would be the density of a glassy surface.

That polystyrene outer layer has a width of$d_{\text{plastic}}=2.0\mathrm{cm}$.

The light speed in oil is obtained from formula $\left(1\right)$

$v_{\mathrm{oil}}=\frac{c}{n_{\mathrm{oil}}}$

Put appropriate data:

$v_{\text{oil}}=\frac{3.0\times {10}^8\mathrm{m}/\mathrm{s}}{1.46}$

$=2.054\times {10}^8\mathrm{m}/\mathrm{s}$

Formula $\left(2\right)$will be used to measure the position travelled for lights through oil layer

$t_{\mathrm{oil}}=\frac{d_{\mathrm{oil}}}{v_{\mathrm{oil}}}$

Put appropriate data:

$t_{\mathrm{oil}}=\frac{\left(5.0\mathrm{cm}\right)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)}{2.054\times {10}^8\mathrm{m}/\mathrm{s}}$

$=2.434\times {10}^{-10}\mathrm{s}$

The light speed in glass is obtained from formula $\left(1\right)$

$v_{\text{glass}}=\frac{c}{n_{\text{glass}}}$

Put appropriate data:

$v_{\text{glass}}=\frac{3.0\times {10}^8\mathrm{m}/\mathrm{s}}{1.50}$

$=2.0\times {10}^8\mathrm{m}/\mathrm{s}$

Formula $\left(2\right)$will be used to measure the position travelled for lights through glass layer

$t_{\text{glass}}=\frac{d_{\text{glass}}}{v_{\text{glass}}}$

Put appropriate data:

$\begin{array}{r}{t}_{\text{glass}}=\frac{\left(1.0\mathrm{cm}\right)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)}{2.0\times {10}^8\mathrm{m}/\mathrm{s}}\\ =5.0\times {10}^{-11}\mathrm{s}\end{array}$

The light speed in polystyrene plastic is obtained from formula $\left(1\right)$

$v_{\text{plastic}}=\frac{c}{n_{\text{plastic}}}$

Put appropriate data:

$\begin{array}{r}{v}_{\text{plastic}}=\frac{3.0\times {10}^8\mathrm{m}/\mathrm{s}}{1.59}\\ =1.886\times {10}^8\mathrm{m}/\mathrm{s}\end{array}$

Formula $\left(2\right)$will be used to measure the position travelled for lights through polystyrene plastic

$t_{\text{plastic}}=\frac{d_{\text{plastic}}}{v_{\text{plastic}}}$

Put appropriate data:

$\begin{array}{r}{t}_{\text{plastic}}=\frac{\left(2.0\mathrm{cm}\right)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)}{1.886\times {10}^8\mathrm{m}/\mathrm{s}}\\ =1.06\times {10}^{-10}\mathrm{s}\end{array}$

The time taken by light which passes through the $8.0cm$sandwich is

$\mathrm{\Delta }t=t_{\text{oil}}+t_{\text{glass}}+t_{\text{plastic}}$

Put appropriate data:

$\begin{array}{r}\mathrm{\Delta }t=\left(2.434\times {10}^{-10}\mathrm{s}\right)+\left(5.0\times {10}^{-11}\mathrm{s}\right)+\left(1.06\times {10}^{-10}\mathrm{s}\right)\end{array}$

$=3.994\times {10}^{-10}\mathrm{s}$

role="math" localid="1649143052302" $=\left(3.994\times {10}^{-10}\mathrm{s}\right)\left(\frac{{10}^9\mathrm{ns}}{1\mathrm{s}}\right)$

$=0.40\mathrm{ns}$