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ArbitrLikvidat [17]
3 years ago
11

Between this and the next assignment, we want to get a better under- standing of how light interacts with the eye. Here are two

questions to get us started, focused on diffraction (i.e., the spreading of light when it passes through a narrow opening). A. To regulate the intensity of light reaching our retinas, our pupils1 change diameter anywhere from 2 mm in bright light to 8 mm in dim light. Find the angular resolution of the eye for 550 nm wavelength light at those extremes. In which light can you see more sharply, dim or bright
Physics
1 answer:
STatiana [176]3 years ago
5 0

Answer:

θ₁ = 3.35 10⁻⁴ rad ,  θ₂ = 8.39 10⁻⁵ rad

Explanation:

This is a diffraction problem for a slit that is described by the expression

       sin θ = m λ

the resolution is obtained from the angle between the central maximum and the first minimum corresponding to m = 1

      sin θ = λ / a

as in these experiments the angle is very small we can approximate the sine to its angle

        θ = λ / a

In this case, the circular openings are explicit, so the system must be solved in polar coordinates, which introduces a numerical constant.

       θ = 1.22 λ / D

where D is the diameter of the opening

 let's apply this expression to our case

indicates that the wavelength is λ = 550 nm = 550 10⁻⁹ m

the case of a lot of light D = 2 mm = 2 10⁻³ m

       θ₁ = 1.22 550 10-9 / 2 10⁻³

       θ₁ = 3.35 10⁻⁴ rad

For the low light case D = 8 mm = 8 10⁻³

      θ₂ = 1.22 550 10-9 / 8 10⁻³

      θ₂ = 8.39 10⁻⁵ rad

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Challenge Problem (Extra Credit) A car with bad shocks has a mass of 1500 kg. Before you go for a drive with three of your frien
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167.354 m

Explanation:

We are given;

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The total speed in the highway; V = 65

mph = 29.058 m/s

The spring's constant can be calculated from the formula;

F = Kx

F is also equal to mg.

Thus;

m_t × g = Kx

K = (m_t × g)/x

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Thus, the period cam be calculated from the formula;

T = 2π√((m_c+t)/k)

T = 2π√(1511/1798.5)

T = 5.759 s

the distance between adjacent bumps is calculated from;

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Distance = 167.354 m

4 0
4 years ago
An atom of element X has one more
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By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 d
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Answer:

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Explanation:

β₁ = sound level of sound at rock concert = 120 dB

β₂ = sound level of sound due to whisper = 21 dB

I₁ = Intensity of sound at rock concert

I₂ = Intensity of sound due to whisper

sound level of sound at rock concert is given as

\beta _{1} = 10 log\left ( \frac{I_{1}}{10^{-12}} \right )

120 = 10 log\left ( \frac{I_{1}}{10^{-12}} \right )

12 = log\left ( \frac{I_{1}}{10^{-12}} \right )               Eq-1

sound level due to whisper is given as

\beta _{2} = 10 log\left ( \frac{I_{2}}{10^{-12}} \right )

21 = 10 log\left ( \frac{I_{2}}{10^{-12}} \right )

2.1 = log\left ( \frac{I_{2}}{10^{-12}} \right )                          Eq-2

subtracting Eq-2 from Eq-1

12 - 2.1 = log\left ( \frac{I_{1}}{10^{-12}} \right ) - log\left ( \frac{I_{2}}{10^{-12}} \right )

9.9 = log\left ( \frac{I_{1}}{I_{2}} \right )

\left ( \frac{I_{1}}{I_{2}} \right ) = 7.94\times 10^{9}

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