Solving Power Problems: Finding d & l

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The discussion focuses on solving two diffraction-related problems using the equation sin(Xmin) = Xmin = 1.22 (lambda/D). For the first problem, with a telescope aperture of 2.32 m and a wavelength of 461 nm, the minimum separation between two stars at a distance of 1022 m is calculated to be 0.248 mm. The second problem involves determining the wavelength of light emitted from a car's taillights, which are 1.14 m apart and observed from 13.8 km away, using the same diffraction principles. The calculations involve rearranging the diffraction equation to find the necessary parameters for both scenarios. The discussion concludes with the application of these principles to derive specific values for minimum separation and wavelength.
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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
 


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)^-
 
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