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arsen [322]
3 years ago
15

The smallness of the critical angle θc for diamond means that light is easily "trapped" within a diamond and eventually emerges

from the many cut faces. This makes a diamond more brilliant than stones with smaller n and larger θc. Traveling inside a diamond, a light ray is incident on the interface between diamond and air. What is the critical angle for total internal reflection? The refraction index for diamond is 2.23 . Answer in uni
Physics
1 answer:
hoa [83]3 years ago
3 0

Answer:

26.6°

Explanation:

refractive index of diamond, n = 2.23

When a ray of light passes from denser medium to the rarer medium and refracts at an angle of 90 degree from the normal of the surface, such angle of incidence in the denser medium is called the critical angle.

By the Snell's law

\frac{Sini}{Sin r }= n

For critical angle, angle of incidence is critical angle, i = θc and angle of refraction, r = 90

So,

Sin θc / Sin 90 = 1 / 2.23

Sin θc = 0.448

θc = 26.6°

Thus, the critical angle is 26.6°.

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Is this right?? please help me. IT IS SOCIOLOGY!!
Westkost [7]

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Yes, it is correct : )

Explanation:

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8 0
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A capacitor is charged to a potential difference of 3 volt it delivers 30% store energy to lamp what is the final potential diff
nexus9112 [7]

Answer:

3.98V

Explanation:

Given

Pontential difference V as 3v

Energy delivered is 30%,

Recall that Enery E=1/2cv^2 from this E=V^2(since Current C is not provided we can assume a value 2)

So E=V^2

E=3^2=9

At full charge E=9,30%of 9,0.3*9=2.7 energy in capacitor is 9-2.7=6.3

But E=V^2

✓E=V

✓6.3=3.98V

4 0
3 years ago
consider a solid sphere and a solid disk wiht the same radius and the same mass. explain why the solid disk has a greater moment
andriy [413]

Answer:

Moment of inertia of the solid sphere:

I

s

=

2

5

M

R

2

.

.

.

.

.

.

.

.

.

.

.

(

1

)

Is=25MR2...........(1)

Here, the mass of the sphere is

M

M

4 0
3 years ago
A silver sphere with radius 1.3611 cm at 23.0°C must slip through a brass ring that has an internal radius of 1.3590 cm at the s
Readme [11.4K]

Answer:

The temperature must the ring be heated so that the sphere can just slip through is 106.165 °C.

Explanation:

For brass:

Radius = 1.3590 cm

Initial temperature = 23.0 °C

The sphere of radius 1.3611 cm must have to slip through the brass. Thus, on heating the brass must have to attain radius of 1.3611 cm

So,

Δ r = 1.3611 cm - 1.3590 cm = 0.0021 cm

<u>The linear thermal expansion coefficient of a metal is the ratio of the change in the length per 1 degree temperature to its length.</u>

<u>Thermal expansion for brass = 19×10⁻⁶ °C⁻¹</u>

Thus,

\alpha=\frac {\Delta r}{r\times \Delta T}

Also,

\Delta T=T_{final}-T_{Initial}

So,

19\times 10^{-6}=\frac {0.0021}{1.3290\times (T_{final}-23.0)}

Solving for final temperature as:

(T_{final}-23.0)=\frac {0.0021}{1.3290\times 9\times 10^{-6}}

<u>Final temperature = 106.165 °C</u>

5 0
3 years ago
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