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vitfil [10]
2 years 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]2 years 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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Explanation:

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We now calculate thus;

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Answer:

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Explanation:

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g = acceleration due to gravity 9.81 m/s2

h = depth of water 11000 m

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Therefore

F = 1025 x 9.81 x 11000 x 0.246

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g In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period
NeTakaya

The question is incomplete. The complete question is :

In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period of the oscillations is 1.2 microseconds (1.2*10^-6). What is Q?

Solution :

The underdamped RLC circuit

$v_{t} = ve^{-\frac{R}{2L}t} \cos \omega t$

$\omega = \sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}= \frac{2 \pi}{T}$

We know in one time period, v = 2v, at t = T, $v_t = 3.8 v$

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Now, Q value $= \frac{1}{R}\sqrt{\frac{L}{C}}$

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              $\frac{1}{LC}=27.43 \times 10^{12}$

∴ $Q=\sqrt{\left(\frac{1}{457.3 \times 10^3}\right)^2 \times 27.43 \times 10^{12}}$

  $Q=\sqrt{131.166}$

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