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Snezhnost
24 days ago
8

Chromatic aberration comes from the fact that different wavelengths of light travel at different speeds through the material of

the lens; that is, they have different indices of refraction, a property known as dispersion. This means that a lens, in effect, has different focal lengths for different wavelengths of light. Consider a lens made to the following specifications: focal length for red light fred=19.57cm, focal length for blue light fblue=18.87cm.
Part A

Consider an object 5.000cm tall placed a distance 30.00cm from the lens. Assuming that the object reflects both red and blue light, find the ratio yredyblueof the height of the red image to the height of the blue image.

Express your answer numerically to three significant figures.
Physics
1 answer:
Yuliya22 [3.3K]24 days ago
5 0

Response:

 y_red / y_blue = 1.11

Clarification:

To determine the image for each wavelength, we'll utilize the lens maker's equation

         1 /f = 1 /o + 1 /i

Where f signifies the focal length, o represents the object distance, and i indicates the image distance

For red light

           1 / i = 1 / f - 1 / o

           1 / i_red = 1 / f_red - 1 / o

           1 / i_red = 1 / 19.57 - 1/30

           1 / i_red = 1.776 10-2

           i_red = 56.29 cm

For blue light

            1 / i_blue = 1 / f_blue - 1 / o

            1 / i_blue = 1 / 18.87 - 1/30

            1 / i_blue = 1.966 10-2

            i_blue = 50.863 cm

Next, we will compute the magnification ratio

             m = y ’/ h = - i / o

             y ’= - h i / o

For red light

            y_red ’= - 5 56.29 / 30

            y_red ’= - 9.3816 cm

For blue light

            y_blue ’= 5 50.863 / 30

            y_blue ’= - 8.47716 cm

The ratio of the heights of both images is

            y_red ’/ y_blue’ = 9.3816 / 8.47716

            y_red / y_blue = 1.107

            y_red / y_blue = 1.11

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An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte
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Answer:

35.79 meters

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v_{arrow} * t_{arrow} = d\\ \\40 \frac{m}{s} * t_{arrow} = d

Right after this, the arrow produces a muffled noise, traveling the same distance d at a speed of 340 m/s in time t_{sound}. Thus, we can derive:

v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d.

Consequently, the sound reaches the archer, precisely 1 second post-firing the bow, resulting in:

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Using this relationship in the distance formula for sound allows us to write:

340 \frac{m}{s} * t_{sound} = d \\ \\ 340 \frac{m}{s} * (1 s- t_{arrow}) = d.

Substituting the value of d from the first equation yields:

40 \frac{m}{s} * t_{arrow} = d \\ 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * (1 s- t_{arrow}).

Now, after some calculations, we can proceed further:

40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * 1 s - 340 \frac{m}{s} * t_{arrow} \\ \\ 40 \frac{m}{s} * t_{arrow} + 340 \frac{m}{s} * t_{arrow} = 340 m \\ \\ 380 \frac{m}{s} * t_{arrow} = 340 m \\ \\ t_{arrow} = \frac{340 m}{380 \frac{m}{s}} \\ \\ t_{arrow} = 0.8947 s.

Finally, the value is inserted into the initial equation:

40 \frac{m}{s} * t_{arrow} = d

40 \frac{m}{s} * 340/380 s = 35,79 s = d

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