An optical fiber consists of an outer "cladding" layer and an inner core with a slightly higher index of refraction. light rays entering the core are trapped inside by total internal reflection and forced to travel along the fiber (see the following figure). suppose the cladding has an index of refraction of 1.46 and the core has an index of refraction of 1.48.calculate the largest angle between a light ray and the longitudinal axis of the fiber for which the ray will be totally internally reflected at the core/cladding boundary.

Answer

given,

refractive index of cladding = 1.46

refractive index of inner core = 1.48

largest angle between light ray = ?

critical angle,

According to Snell's law

assuming angle of incident be equal to 50°

then Largest angle

        =50° - 14° = 36°

c. rb

explanation:

the atomic radius of a chemical element is a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding shells of electrons.

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz