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Vinvika [58]
2 years ago
8

Most of the bending of light in the eye is done at the air-cornea interface. By what angle (θcornea) is the beam of light shown

in Figure deviated as it passes from air to the cornea if the incident angle is θi = 23.6°? The refractive index of air is nair = 1.00, the refractive index of the cornea is ncornea = 1.38. Explain why θcornea< θi?
Physics
1 answer:
Paladinen [302]2 years ago
4 0

The angle (θcornea)  when light passes from air to cornea is 16.86°

<h3>What is Snell's law?</h3>

It states that the ratio of sine of angle of incidence and angle of refraction is equal to the refractive index of second medium to the  first medium.

sini/sinr =n₂ / n₁

Most of the bending of light in the eye is done at the air-cornea interface. The beam of light deviated as it passes from air to the cornea if the incident angle is θi = 23.6°.

Given the angle of incidence i = 23.6°, refractive index of air n₁ =1, refractive index of cornea n₂ = 1.38, then the angle of refraction at cornea is

sinr = sini x (n₁/n₂)

Plug the values, we get

sinr = sin23.6 x (1/1.38)

sinr = 16.86°

The angle of refraction is less than angle of incidence due to refraction.

Thus, the angle (θcornea)  when light passes from air to cornea is 16.86°

Learn more about Snell's law.

brainly.com/question/10112549

#SPJ1

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A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f
VashaNatasha [74]

Answer:

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34  

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 =  X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + (\frac{L-y}{2}) ) g sinФ - F2 x (\frac{L}{2}) x gsinФ = 0

plug in the  equations of F1 and F2 into the above equation and after simplification, we get:

(L^{2} - y^{2} ) . w = L^{2} . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

(L^{2} - y^{2} ) . w = L^{2} . 5/9 x w

Hence,

(L^{2} - y^{2} ) = \frac{5L^{2} }{9}

\frac{L^{2} - y^{2}  }{L^{2} } = \frac{5}{9}

Taking L^{2} common and solving for \frac{y}{L}, we will get

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34

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Which of The following is the best example of water changing from a liquid to gas
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Answer:

There are no examples but this should be evaporation

Explanation:

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A dolphin can swim at a constant speed of 12.5 m/s. How
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Answer:

\boxed {\tt 3.6 \ seconds}

Explanation:

Time can be found by dividing the distance by the speed.

t=\frac{d}{s}

The distance is 45 meters and the speed is 12.5 meters per second.

d= 45 \ m \\s= 12.5 \ m/s

t=\frac{45 \ m}{12.5 \ m/s}

Divide. Note that the meters, or "m" will cancel each other out.

t=\frac{45 }{12.5 \ s}

t=3.6 \ s

It will take the dolphin 3.6 seconds to swim a distance of 45 meters are 12.5 meters per second.

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Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point. F⃗ 1F→1F_1_vec has a magnitude of 9.20 NN and is directed at an a
BARSIC [14]

Answer:

The x component of the resultant force is -7.27N.

Explanation:

To obtain the x component of the resultant force, first we have to know the x components of the other forces. To do this, we just have to do some trigonometry:

|F_{1x}|=|F_1|\cos\theta_1=9.20N\cos62.0\°=4.31N \\|F_{2x}|=|F_2|\cos\theta_2=5.00N\cos53.6\°=2.96N

Since both vectors are in the left side of the y-axis, they have a negative x component. So:

F_{1x}=-4.31N;\\F_{2x}=-2.96N

Finally, we sum both components to obtain the component of the resultant force:

F_{Rx}=-4.31N-2.96N=-7.27N

In words, the x component of the resultant force is -7.27N.

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3 years ago
An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu
ExtremeBDS [4]

Explanation:

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Since, the electron is travelling downwards it means that it looses the potential energy.

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