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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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Answer:

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fredd [130]

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8 0
3 years ago
Which SI unit is the correctly abbreviated for describing the mass on an object.
djverab [1.8K]
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In short, Your Answer would be Option C

Hope this helps!
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3 years ago
An automobile starter motor has an equivalent resistance of 0.055 Ï and is supplied by a 12.0 v battery which has a 0.0305 Ï int
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8 0
3 years ago
When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a
sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = \frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}

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or,

  = \frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}

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(b)

As we know,

⇒ Vq=\frac{hc}{\lambda}-\Phi_0

By substituting the values, we get

⇒ 1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0

⇒                       \Phi_0=2.3\times 10^{-19} \ J

or,

⇒                            =\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}

⇒                            =1.4375 \ eV

5 0
3 years ago
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