Answer to Question #108867 in Optics for princess

Question

A layer of water covers an unknown transparent material X. At the angles shown in the figure below, a ray of light travels from material X to water which has an index of refraction n = 1.33.
(a) Show your calculation of the index of refraction of the unknown material X.
Eventually the light ray will exit water into air, which has an index of refraction n = 1.00.
(b) Show your calculation of the angle at which the light ray will emerge into air.

Answer 1

As per the given question,

The refractive index of the material $(\mu)=1.33$

Diagram is missing in the question,

i) Refractive index of water $(\mu_w)=1$

so shift in the ray =$ti(1-\dfrac{\mu_w}{\mu})$

angle of incidence is not given

$=t i (1-\dfrac{1}{1.33})=\dfrac{1}{3}ti$

where t is the thickness and i is the angle of incidence.

ii)For the first layer, applying the snell's law,

$\dfrac{\sin \theta_1}{\sin \theta_2}=\dfrac{3}{4}$

for the second layer

$\dfrac{\sin \theta _2}{\sin\theta_3}=\dfrac{\mu_{air}}{1}$

Hence from above,

$\sin\theta_3=\dfrac{4}{3}\sin\theta_1$