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Levart [38]
3 years ago
9

To regulate the intensity of light reaching our retinas, our pupils1 change diameter anywhere from 2 mm in bright light to 8 1 T

he pupil of the eye is the circular mm in dim light. Find the angular resolution of the eye for 550 nm opening through which light enters. wavelength light at those extremes. In which light can you see more sharply, dim or bright
Physics
1 answer:
Ronch [10]3 years ago
4 0

Correct question is;

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?

Answer:

We'll see more sharply in dim light

Explanation:

If we consider diffraction through a circular aperture, then angular resolution is given by;

θ = 1.22λ/D

where:

θ is the angular resolution (radians) λ is the wavelength of light

D is the diameter of the lens' aperture.

Thus,

at diameter = 2mm = 2 x 10^(-3) m = 2 x 10^(6) nm

θ = (1.22 * 550)/(2 x 10^(6))

θ = 335.5 x 10^(-6) radians

Now, we need to convert this to arc seconds.

Thus;

1 arc second = 4.85 x 10^(-6) radians

So,θ = 335.5 x 10^(-6) radians = [335.5 x 10^(-6)]/[4.85 x 10^(-6)]

= 69.18 arc seconds

at diameter = 8mm = 8 x 10^(-3) m = 8 x 10^(6) nm

θ = (1.22 * 550)/(8 x 10^(6))

θ = 83.875 x 10^(-6) radians

Now, we need to convert this to arc seconds.

Thus;

1 arc second = 4.85 x 10^(-6) radians

So,θ = 83.875 x 10^(-6) radians = [83.875 x 10^(-6)]/[4.85 x 10^(-6)]

= 17.3 arc seconds

From the values of angular resolution gotten, we see that sharpness of image increases with increasing angular resolution. Thus, objects are sharper in dim light.

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1)  \rm q_2 is<u> positive.</u>

<u></u>

2) \rm q_2=4.56\times 10^{-10}\ C.

Explanation:

<h2><u>Part 1:</u></h2>

<u></u>

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., \rm q_2 is <u>positive.</u>

<u></u>

<h2><u>Part 2:</u></h2>

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<u>Given:</u>

  • Mass of the balloon, m = 0.00275 kg.
  • Charge on the balloon, \rm q_1 = -3.50\times 10^{-8}\ C.
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In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

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The magnitude of the electrostatic force on the balloon due to the rod is given by

\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.

\rm \dfrac{1}{4\pi \epsilon_o} is the Coulomb's constant.

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\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.

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Hopr it helps :)
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