The index of refraction of quartz is anisotropic. Suppose that visible light is passing from one grain to another of different crystallographic orientation and at normal incidence to the grain boundary. Calculate the reflectivity at the boundary if the indices of refraction for the two grains are 1.545 and 1.555 in the direction of light propagation. Enter your answer in accordance to the question statement

1.04*10⁻⁵

Explanation:

light wave do showcase some behaviors whenever there is encounters with the end of the medium, some of the behaviors are - reflection, refraction, as well as diffraction. When visible light wave strikes a boundary that exist two different media, a portion of the energy will be transmitted into the new medium and some reflected.

Reflection of a light wave can be regarded as bouncing off of light wave from boundary. refraction on other hand is bending of the path of a light wave.

We were to calculate the reflectivity at the boundary,

reflectivity at the boundary can be calculated using the expression below

Reflectivity= (n₂ - n₁)² /(n₂ + n₁ ) ²

where

n₁= Indices of refraction at first grain= 1.545

n₂= Indices of refraction at second grain=

1.555

(1.555 - 1.545)² / (1.555 - 1.545)²

=(0.01)²/ (3.1)²

= 0.0001/ 9.61

= 1.04*10⁻⁵

Hence, the reflectivity at the boundary if the indices of refraction for the two grains are 1.545 and 1.555 in the direction of light propagation is 1.04*10⁻⁵

colorectal cancer may also develop from areas of abnormal cells in the lining of the colon or rectum. this area of abnormal cells is called dysplasia and is more commonly seen in people with certain inflammatory diseases of the bowel such as crohn's disease or ulcerative colitis.

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6