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Alinara [238K]
3 years ago
6

Using Rayleigh's criterion, what is the smallest separation between two pointlike objects that a person could clearly resolve at

his near point of 20.5 cm with a pupil diameter of 2.15 mm ? Assume that the light has a wavelength of 550 nm . (To get a reasonable estimate without having to go through complicated calculations, we'll ignore the effect of the fluid in the eye.)
Physics
1 answer:
atroni [7]3 years ago
7 0

 Answer:

y = 52.44 10⁻⁶  m

Explanation:

It is Rayleigh's principle that two points are resolved if the maximum of the diffraction pattern of one matches the minimum the diffraction pattern of the other

Based on this principle we must find the angle of the first minimum of the diffraction expression

         a sin θ= m λ

The first minimum occurs for m = 1

       sin θ =  λ  / a

Now let's use trigonometry the object is a distance L = 0.205 m

        tan θ = y / L

Since the angles are very small, let's approximate

        tan θ = sin θ/cos θ = sin  θ

        sin θ = y / L

We substitute in the diffraction equation

         y / L =  λ  / a

         y =  λ  L / a

Let's calculate

        y = 550 10⁻⁹ 0.205 / 2.15 10⁻³

        y = 52.44 10⁻⁶ m

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Explanation:

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Now, we will differentiate the equation with respect to t as follows.

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(b)  Wavelength of the wave is given as follows.

               \lambda = \frac{2\pi}{k}

              \lambda = (frac{2\pi}{\frac{\pi}{11})


              \lambda = 22 m

The period of the wave is

             T = \frac{2 \pi}{\omega}

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                    v = \frac{\lambda}{T}

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