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vitfil [10]
1 year ago
14

By how much do the critical angles for red (660 nm) and violet (410 nm) light differ in a diamond surrounded by air?

Physics
1 answer:
Stella [2.4K]1 year ago
3 0

the critical angle for red light in diamond is 24.51°

critical angle for violet light in diamond is 24.01°.

given:

wavelength of red light λr =660nm

wavelength of violet light λv =410nm

refractive index for air=1

as the refractive index of red and violet colour in diamond is

Refractive index of red color nr= 2.407

Refractive index of violet color nv= 2.451

as it is mentioned that angle is critical angle,thus θr =90°

so,from Snell's law

n1sinθc=n2 sin 90°

Here,

n1 is the refractive index for incident medium

n2 is the refractive index for refractive medium

θc is the critical angle after which light starts reflecting internally

The critical angle for red light is here,n1=2.407, n2=1

n1sinθc=n2 sin 90°

2.407×sin θc=1×1

sin θc =(1/2.407)

θc=sin-1 0.415

θc=24.51°

Thus,the critical angle for red light in diamond is 24.51°

The critical angle for violet light is,

here,n1=2.451,n2=1

n1sinθc=n2 sin 90°

2.407×sin θc=1×1

sin θc= (1/2.451)

θc=sin-1 0.407

=24.01°

Thus,the critical angle for violet light in diamond is 24.01°.

learn more about Snell's law from here: brainly.com/question/2273464

#SPJ4

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You're driving a vehicle of mass 1350 kg and you need to make a turn on a flat road. The radius of curvature of the turn is 71 m
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Answer:

v=12.65\ m.s^{-1}

Explanation:

Given:

  • mass of vehicle, m=1350\ kg
  • radius of curvature, r=71\ m
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<u>During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:</u>

m.\frac{v^2}{r} =\mu.N

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\mu= coefficient of friction

N= normal reaction force due to weight of the car

v= velocity of the car

1350\times \frac{v^2}{71} =0.23\times (1350\times 9.8)

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Which is not an example of an external force acting on an object? (1 point)
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A 60 g golf ball is dropped from a level of 2 m high. It rebounds to 1.5 m. How much energy is lost? Group of answer choices 0.5
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Answer: A 60 g golf ball is dropped from a level of 2 m high. It rebounds to 1.5 m. Energy loss will be 0.29J

Explanation: To find the correct answer, we have to know more about the Gravitational potential energy.

<h3>What is gravitational potential energy?</h3>
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  • The gravitational potential energy of a body at a height h with respect to the height h will be,

                                          U=mgh

  • Expression for gravitational potential energy loss will be,

                                        E=U_i-U_f

<h3>How to solve the problem?</h3>
  • The total energy before the ball dropped will be,

                 U_i=mgh_i=60*10^-3kg*9.8m/s^2*2m=1.176 J

  • The total energy after when the ball rebounds to 1.5m will be,

                 U_f=mgh_f=60*10^-3kg*9.8m/s^2*1.5m=0.882J

  • The total energy loss will be,

                E=1.176-0.882=0.294J

Thus, we can conclude that, the energy loss will be,0.294J.

Learn more about the gravitational potential energy here:

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