Solving Power Problems: Finding d & l

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In summary, the conversation discusses two separate problems related to resolving power. The first problem involves calculating the minimum separation between two stars in order to distinguish them as separate objects, while the second problem involves determining the wavelength of light emitted by a car's taillights based on the observed diffraction pattern. The solutions to both problems are not provided in the conversation.
  • #1
PixelHarmony
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Easy Resolving Power Problem

Yet I can't get the answers. Can someone please help me?

sin(Xmin) = Xmin = 1.22 (lambda/D)

(a) Two stars are photographed utilizing a telescope with a circular aperture of diameter of 2.32 m and light with a wavelength of 461 nm. If both stars are 1022 m from us, what is their minimum separation so that we can recognize them as two stars (instead of just one)?

d = ? m*****

(b) A car passes you on the highway and you notice the taillights of the car are 1.14 m apart. Assume that the pupils of your eyes have a diameter of 6.7 mm and index of refraction of 1.36. Given that the car is 13.8 km away when the taillights appear to merge into a single spot of light because of the effects of diffraction, what wavelength of light does the car emit from its taillights (what would the wavelength be in vacuum)?

l = ? nm*****
 
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nevermind i got them
 
  • #3


To solve these problems, we need to use the equation sin(Xmin) = Xmin = 1.22 (lambda/D), where Xmin is the minimum angular separation between the two objects, lambda is the wavelength of light, and D is the diameter of the aperture. We can rearrange this equation to solve for D by dividing both sides by 1.22 and multiplying by lambda. This gives us D = lambda / 1.22sin(Xmin).

(a) For the first problem, we are given the diameter of the aperture (D = 2.32 m) and the wavelength of light (lambda = 461 nm = 4.61 x 10^-7 m). We are also given the distance to the stars (r = 1022 m). Plugging these values into the equation, we get:

Xmin = 1.22 (4.61 x 10^-7 m) / (2.32 m) = 2.43 x 10^-7 rad

To find the minimum separation between the stars, we can use the formula s = rXmin, where s is the separation between the stars and r is the distance to the stars. Plugging in the values, we get:

s = (1022 m)(2.43 x 10^-7 rad) = 2.48 x 10^-4 m = 0.248 mm

So the minimum separation between the stars should be 0.248 mm for us to be able to recognize them as two separate objects.

(b) For the second problem, we are given the distance to the car (r = 13.8 km = 1.38 x 10^4 m) and the diameter of our pupils (D = 6.7 mm = 6.7 x 10^-3 m). We are also given the index of refraction of our eyes (n = 1.36). We can use the same equation as before, but this time we need to solve for lambda. Rearranging the equation, we get:

lambda = D1.22sin(Xmin)

To find Xmin, we can use the formula Xmin = s/r, where s is the separation between the taillights and r is the distance to the car. Plugging in the values, we get:

Xmin = (1.14 m)(1.38 x 10^4 m)^-
 

1. How do you determine the value of d and l in a power problem?

In order to solve for the values of d (distance) and l (time), you need to have at least two known values from the power equation. This could be the power output, force, or velocity. Once you have these values, you can use the equations P = F x v and P = W/t to solve for d and l respectively.

2. Can you solve for d and l if only one value is known?

No, in order to solve for both d and l, you need to have at least two known values from the power equation. If you only have one known value, you can only solve for one variable.

3. How does the unit of measurement affect the calculation of d and l?

The unit of measurement for the values used in the power equation must be consistent. For example, if the power output is measured in watts, the force should be measured in newtons and the velocity in meters per second. Using inconsistent units will result in incorrect calculations.

4. What are some common mistakes to avoid when solving power problems?

One common mistake is using the wrong formula. Make sure to use the correct formula depending on the known values given in the problem. Another mistake is not converting units correctly, which can lead to incorrect calculations. Lastly, be careful with the order of operations when solving for d and l, as this can also impact the final result.

5. Are there any real-life applications for solving power problems?

Yes, power problems can be found in a variety of fields including physics, engineering, and sports. For example, calculating the power output of a car engine or the power needed to lift weights in the gym. Understanding how to solve power problems can also help with understanding the efficiency of machines and systems in everyday life.

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