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DIA [1.3K]
4 years ago
7

The shortest wavelengths that we can see are experienced as what colors?

Physics
1 answer:
astra-53 [7]4 years ago
6 0
If you think about the colors of the rainbow red, orange, yellow, green, blue, indigo, and violet.  That list of colors shows you the spectrum of visible light and the further right (towards violet) you go the more energy the waves have while the more left (towards red) you go the less energy the waves have.  wavelength and energy are inversely proportional meaning that as energy goes up, wavelength goes down.  therefore the more right (towards violet) you move in the spectrum, the smaller the wavelengths get.
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Explanation:

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3 years ago
What is a descriptions of an animal cell's nucleus?
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Can two similar charges attracts each other ? ​
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5 0
3 years ago
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Light of wavelength 630 nm is incident on a long, narrow slit. Determine the angular deflection of the first diffraction minimum
asambeis [7]

Answer:

a) 1.8°

b) 0.18°

c) 0.018°

Explanation:

Wavelength (λ) = 630nm = 630 *10^-9m

The equation that describes the angular deflection of a dark band is

Wsin(βm) = mλ

w = width of the single slit

λ = wavelength of the light

βm = angular deflection of the mth dark band.

a) In order to get the angular deflection of the first dark band for a slit with 0.02mm width, substitute w = 0.02mm = 0.02*10^-3 , m = 1 , λ= 630*10^-9

0.02*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.02*10^-3

Sin(β1) = 0.0315

β1 = Sin^-1(0.0315)

= 1.8°

b) substitute w = 0.2mm = 0.2*10^-3 , m = 1 , λ= 630*10^-9

0.2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.2*10^-3

Sin(β1) = 0.00315

β1 = Sin^-1(0.00315)

= 0.18°

c) substitute w = 2mm = 2*10^-3 , m = 1 , λ= 630*10^-9

2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 2*10^-3

Sin(β1) = 3.15*10^-4

β1 = Sin^-1(3.15*10^-4)

= 0.018°

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3 years ago
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Inessa05 [86]
C I think not totally sure
4 0
3 years ago
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