Answer:
is reflected back into the region of higher index
Explanation:
Total internal reflection is a phenomenon that occurs when all the light passing from a region of higher index of refraction to a region of lower index is reflected back into the region of higher index.
According to Snell's law, refraction of ligth is described by the equation
where
n1 is the refractive index of the first medium
n2 is the refractive index of the second medium
is the angle of incidence (in the first medium)
is the angle of refraction (in the second medium)
Let's now consider a situation in which
so light is moving from a medium with higher index to a medium with lower index. We can re-write the equation as
Where 
is a number greater than 1. This means that above a certain value of the angle of incidence , the term on the right can become greater than 1. So this would mean
But this is not possible (the sine cannot be larger than 1), so no refraction occurs in this case, and all the light is reflected back into the initial medium (total internal reflection). The value of the angle of incidence above which this phenomen occurs is called critical angle, and it is given by
If Resistors are in series= The equivalent is the sum.
E.g R1 and R2 in series, R = R1 + R2.
If in Parallel, equivalent is Product/sum.
E.g If R1 and R2 in parallel, R = (R1*R2)/(R1+R2)
1.) 60 is parallel with 40 and both are then in series with 20.
60//40 = (60*40)/(60+40) = 2400/100 = 24
Now the 24 is in series with the 20
R = 24 + 20 = 44 ohms.
2.) 80 is in series with 40 and both are then in parallel with 40.
Solving the series, R = 80 + 40 =120.
Parallel: 120//40 = (120*40)/(120+40) = 4800/160 = 30
Equivalent Resistance = 30 ohms.
3.) 100 is in parallel with 100 and both are then in series with the parallel of 50 and 50.
The 1st parallel = (100*100)/(100+100) = 10000/200 = 50
The 2nd parallel = (50*50)/(50+50) = 2500/100 = 25.
Solving the series = 50 + 25 =75 ohms.
Cheers.
Answer:
q=6.22*10^-10C
Explanation:
Two large metal plates of area 0.88 m2 face each other, 4.8 cm apart, with equal charge magnitudes but opposite signs. The field magnitude E between them (neglect fringing) is 80 N/C. Find |q|
E=α/∈, electric field within the plate
α=q/A
A=area of the plate
∈=is the permittivity
substituting , we have
The field magnitude E between them (neglect fringing)
E=q/A∈
q=EA∈
q=0.88*80*8.84*10^-12
q=6.22*10^-10C
<em>Answer:</em>
<em>Answer:It is due to tyndall effect that the path of a light ray passing through a suspension is visible. The particles in suspension scatter the beam of light making it clearly visible.</em>
