Answer to Question #287124 in Optics for kelly

Question

. A container with glass walls has oil inside. The total length,  of the containers is 14 cm. The walls of the container have a thickness () of 25 mm and a refractive index of 1.58. The container is surrounded by air (n = 1), and the oil inside the container has a refractive index of 1.42. The angle of incidence,  is 30º. Calculate the lateral shift, x, for this container.. A container with glass walls has oil inside. The total length,  of the containers is 14 cm. The walls of the container have a thickness () of 25 mm and a refractive index of 1.58. The container is surrounded by air (n = 1), and the oil inside the container has a refractive index of 1.42. The angle of incidence,  is 30º. Calculate the lateral shift, x, for this container.  

Answer 1

Solution:

Thickness of walls of container $\left(t_{1}\right)=2.5 \mathrm{~cm}$

So, Thickness of oil (which is inside the container)

$\begin{aligned} t_{2}=l-2 t_{1} &=14 \mathrm{~cm}-2 \times 2.5 \mathrm{cm} \\ &=9 \mathrm{~cm} \end{aligned}$

Now,

For first lateral shift of $\Delta x_{1}$ (when lignt travels from air into glass wall).

$\begin{aligned} i=30^{\circ}, & n_{1}=1, \quad n_{2}=1.58 \\ t=t_{1}=2.5 \mathrm{~cm} & \\ \text { so, } & 1 \times \sin 30^{\circ}=1.58 \times \sin r_{1} \\ & \sin r_{1}=\frac{0.5}{1.58}=0.32 \\ & r_{1}=18.45^{\circ} \end{aligned}$

$\text { So, } \begin{aligned} \Delta x_{1} &=2.5 \mathrm{~cm} \times \frac{\sin \left(30^{\circ}-18.45^{\circ}\right)}{\cos (18.450)} \\ &=0.528 \mathrm{~cm} \end{aligned}$

For second latent shift of $\Delta x_{2}$ (when light travels from wall into oil)

$\begin{aligned} i_{2}=r_{1}=18.45\degree \quad n_{2}=1.58 \quad n_{3}=1.42 \\ t=t_{2}=9 \mathrm{~cm} \\ \text { so, } \quad 1.58 \times \sin (18.45)=1.42 \times \sin r_{2} \\ \Rightarrow \quad r_{2}=20.62^{\circ} \\ \text { so, } \begin{aligned} \Delta x_{2} &=9 \mathrm{~cm} \times \frac{\sin (18.45-20.62)}{\cos (20.62)} \\ &=-0.364 \mathrm{~cm} \end{aligned} \end{aligned}$

$\Rightarrow negative\ \Delta x_{2}$ means light is traveling from denser to rarer medium $\left(n_{2}>n_{3}\right)$ . So, it bends towards normal.

For third lateral shift of $\Delta x_{3}$ (when light trand from oil into wall of container)

$\begin{aligned} &i_{3}=r_{2}=20.62^{\circ} \quad n_{4}=n_{2}=1.58 \quad n_{3}=1.42 \\ &t=t_{1}=2.5 \mathrm{~cm} \\ &\text { so, } \quad 1.42 \times \sin (20.62)=1.58 \times \sin r_{3} \\ &\Rightarrow r_{3}=18.45^{\circ} \end{aligned}$

So,

$\begin{aligned} \Delta x_{3} &=2.5 \mathrm{~cm} \times \frac{\sin (20.62-18.45)}{\cos 18.45} \\ &=0.1 \mathrm{~cm} \end{aligned}$

Total lateral shift

$\begin{aligned} \Delta x &=\Delta x_{1}+\Delta x_{2}+\Delta x_{3} \\ &=0.528 \mathrm{~cm}-0.364 \mathrm{~cm}+0.1 \mathrm{~cm} \\ &=0.264\mathrm{~cm} \end{aligned}$