Hey mates. I need this answered ASAP!

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

Ronch [10]

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.

4 0

4 years ago

If an equivalent (eq) is defined as a mole of charge, what is the concentration of ca2 in blood plasma in eq if its concentratio

Andrei [34K]

The concentration of ca2 in blood plasma is comparable if it is 0.0080, according to the supplied statement.

<h3>How do you quantify concentration?</h3>

Subtract the solute's mass from the total volume of the mixture. Calculate C = m/V, where m is the solute's mass and V seems to be the total volume of such solution. Divide the figures you found for mass and volume to find approximate concentration for your solution.

<h3>What is the definition of concentration in the solution?</h3>

A solution's concentration is a measure of how much solute has been dissolved in a specific amount of the solvent or solution. A concentrated solution contains a significant amount or dissolved solute. A dilute solution is one with a little amount of dissolved solute.

<h3>According to the given data:</h3>

The equivalent weight of detail is the mass that combines with someone or displaces 1.008 grams of molecular hydrogen, or eight.

35.5 grams of chlorine or 0 grams of oxygen.

Those figures are the atomic weight divided by that of the normal valence, which is sixteen in the case of oxygen.

One Ca2+ ion has a charge of +2.Calcium ion concentration:

= 0.0040M ⇒ Ca 2+ Concentration in blood plasma

= 0.0040 × 2⇒ 0.0080 equivalent Ca2

To know more about concentration visit: brainly.com/question/10725862

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7 0

2 years ago

The rate at which an object moves

sergejj [24]

The rate at which an object moves is called speed or velocity depending if the direction does or doesnt matter

3 0

3 years ago

Read 2 more answers

A force of 1.5 × 102 N is exerted on a charge of 1.4 × 10–7 C that is traveling at an angle of 75° to a magnetic field.

Marysya12 [62]

Option (B) is correct.The magnetic field strength=8.5 x 10² T

Explanation:

the magnetic force Fm is given by

Fm= q V B sinθ

q= charge=1.4 x 10⁻⁷ C

v= velocity= 1.3 x 10⁶ m/s

B= magnetic field strength

Fm= magnetic force= 1.5 x 10² N

θ=75°

so 1.5 x 10²=(1.4 x 10⁻⁷) (1.3 x 10⁶ ) (B) sin75

B=8.5 x 10² T

6 0

4 years ago

Read 2 more answers

A sound wave traveling at 343 m/s is emitted by the foghorn of a tugboat. an echo is heard 2.80 s later. how far away is the ref

Nostrana [21]

The echo is heard 2.80 s later, this means this is the time the sound takes to travel to the reflecting object and then back to us. So, during this time, the sound wave has covered the distance L between us and the object twice:
$S=2L$
The speed of the sound wave is: $v=343 m/s$ , and since it is moving by uniform motion, we can find the distance covered by the wave using
$S=vt=(343 m/s)(2.80 s)=960 m$
And we said this corresponds to twice the distance between us and the reflecting object, so:
$L= \frac{S}{2} = \frac{960 m}{2} =480 m$
so, the object is 480 meters away.

3 0

4 years ago