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Umnica [9.8K]
3 years ago
13

At the interface of two transparent media, light ray experiences both refraction and reflection. Does the angle of reflection de

pend on the angle of refraction? Can you demonstrate the effect of total internal reflection using the rectangular block? Explain your answer and support your answer with a drawing. In which case the shift of the light beam passing through the transparent block equals zero?

Physics
1 answer:
juin [17]3 years ago
6 0

Answer:

No, the angle of reflection is independent on the angle of refraction. The total internal reflection is when the angle of incidence is more than the critical angle.

Explanation:

No, the angle of reflection is independent on the angle of refraction. The angle of reflection is only affected by the angle of incidence.

The total internal reflection is when the angle of incidence (I) is more than the critical angle (C). When the light ray pass through the rectangular block, with time, a critical angle will be formed when the refracted ray is exactly 90 degrees to the normal. As soon as the angle of incidence is more than the critical angle, a total internal reflection is formed (as shown in the attached figure).

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The solid rod shown has a radius of 0.75 in. if it is subjected to the force the 500lb determine the max normal stress developed
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On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wi
Lostsunrise [7]

Answer:

0.06\Omega/m

Explanation:

Firstly, when you measure the voltage across the battery, you get the emf,

E = 13.0 V

In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.

Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire (I=\frac{V}{R}). But this is not the case.

Let the resistance of the ammeter be r

Hence, using Ohm's law we get the following 2 equations:

\frac{13}{20z+r} =7.6   .......(1)

\frac{13}{40z+r} =4.5     ......(2)

Substituting the value of r from (2) in (1), we have,

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putting the value of z in either (1) or (2) gives us, r = 0.5325 \Omega

3 0
3 years ago
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