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3 years ago

5

Calculate the diffraction limit of the human eye, assuming a wide-open pupil so that your eye acts like a lens with diameter 0.8

centimeter, for visible light of 500-nanometer wavelength.

1 answer:

guapka [62]3 years ago

6 0

Hi, thank you for posting your question here at Brainly.

This problem could be solved using this equation:

Diffraction limit = 1.22*wavelength/diameter

diameter = 0.8 cm = 0.008 m
wavelength = 500E-9 m

Diffraction limit = 1.22(500E-9)/0.008
Diffraction limit = 0.00007625

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John says that the value of the function cos[ω(t + T) + ϕ], obtained one period T after time t, is greater than cos(ωt + ϕ) by 2

Svetllana [295]

Answer:

No one is right

Explanation:

John Case:

The function $cos(\omega t +\phi)$ is defined between -1 and 1, So it is not possible obtain a value $2\pi$ greater.

In addition, if you move the function cosine a T Value, and T is the Period, the function take the same value due to the cosine is a periodic function.

Larry case:

Is you have $f=1+cos(\omega t +\phi)$ , the domain of this is [0,2].

it is equivalent to adding 1 to the domain of the $f=1+cos(\omega t +\phi)$ , and its mean that the function $f=cos(\omega t +\phi)$ , in general, is not greater than $cos(\omega t +\phi)$ .

3 0

3 years ago

Reading glasses with a power of 1.50 diopters make reading a book comfortable for you when you wear them 1.8 cmcm from your eye.

Vlad1618 [11]

Answer:

The near point is $n =44.8 \ cm$

Explanation:

From the question we are told that

The power is $P = 1.50$

The distance from the eye is $k = 1.8 \ cm$

The distance of the book from the eye is $z = -28 \ cm$

Generally the focal length of the glasses is

$f = \frac{1}{P}$

=> $f = \frac{1}{1.50 }$

=> $f = 0.667 \ m$

=> $f = 66.7 \ cm$

The object distance is evaluated as

$u = z + k$

=> $u = -28 + 1.8$

=> $u = -26.2 \ cm$

The image distance is evaluated from lens formula as

$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$

=> $\frac{1}{v} = \frac{1}{66.7} + \frac{1}{-26.2}$

=> $v=- \frac{1}{0.0232}$

=> $v=- 43 \ cm$

The near point is evaluated as

$n = -v + k$

=> $n =-(-43) + 1.8$

=> $n =44.8 \ cm$

3 0

3 years ago

Three moles of an ideal gas are taken around the eyele acb shown in Fig.For this gas Co-29.2J/mol-K.Process ac is at constant pr

Scorpion4ik [409]

For three moles of an ideal gas are taken around the eye, the total work for the cycle is mathematically given as

Wt=1945.475J

<h3>What is the total work F for the cycle.(R=8.31J/molK)?</h3>

Generally, the equation for work is mathematically given as

Wt=wac+Wc+Wba

Therefore

Wac=Pa(Vc-Va)=nR(Tc-Ta)

Wac=3(8.314*192)

Wac=4788.864J

Wcb=P1v1-p2v2/v-1

Wcb-3*8.34*108/0.4

Wcb=-6734.34J

Wab=0

In conclusion,

Wt=wac+Wc+Wba

Wt=4788.864J+0+(-6734.34J)

Wt=+1945.475J

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2 years ago

You venture out on a cold winter morning to warm up your vehicle. You have layers of cotton/polyester blend clothes on and from

xxMikexx [17]

Answer:

There is a localization of negative charge near the door handle.

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3 years ago

Plane polarized light with intensity I0 is incident on a polarizer. What angle should the principle axis make with respenct to t

Nataliya [291]

Answer:

Q = 47.06 degrees

Explanation:

Given:

- The transmitted intensity I = 0.464 I_o

- Incident Intensity I = I_o

Find:

What angle should the principle axis make with respect to the incident polarization

Solution:

- The relation of transmitted Intensity I to to the incident intensity I_o on a plane paper with its principle axis is given by:

I = I_o * cos^2 (Q)

- Where Q is the angle between the Incident polarized Light and its angle with the principle axis. Hence, Using the relation given above:

Q = cos ^-1 (sqrt (I / I_o))

- Plug the values in:

Q = cos^-1 ( sqrt (0.464))

Q = cos^-1 (0.6811754546)

Q = 47.06 degrees

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4 years ago