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## Why is the sky blue?

The sun emits light at many different wavelengths, including all of the visible wavelengths. Light which is made up of all the visible wavelengths appears white. So what causes the sky to look blue? The atmosphere consists of molecules of different gases as well as tiny dust grains. Light from the sRead more

The sun emits light at many different wavelengths, including all of the visible wavelengths. Light which is made up of all the visible wavelengths appears white. So what

See lesscauses the sky to look blue?

The atmosphere consists of molecules of different gases as well as tiny dust grains.

Light from the sun scatters off the molecules in the air (called Rayleigh scattering). The

chance that the light will scatter off the gas molecules is higher for shorter wavelengths.

The short wavelength blue light is therefore scattered more than the other colours.

At noon, when the light from the sun is coming straight down (see the picture), the

scattered blue light reaches your eyes from all directions and so the sky appears blue.

The other wavelengths do not get scattered much and therefore miss your line of sight

and are not seen. At sunrise or sunset, the direction of the light coming from the sun is

now straight towards your eyes (see the picture). Therefore the scattered blue light can’t

be seen because it is scattered out of your line of sight. The redder colours (oranges

and reds) can now be seen because they are not scattered as much and still fall in your

line of sight.

## I have a ruler which reflects red light and absorbs all other colours of light. What colour does the ruler appear in white light? What primary pigments must have been mixed to make the pigment which gives the ruler its colour?

Step 1 : What is being asked and what are we given? We need to determine the colour of the ruler and the pigments which were mixed to make the colour. Step 2 : An opaque object appears the colour of the light it reflects The ruler reflects red light and absorbs all other colours. Therefore the rulerRead more

See lessStep 1 : What is being asked and what are we given?We need to determine the colour of the ruler and the pigments which were mixed to make the colour.

Step 2 : An opaque object appears the colour of the light it reflectsThe ruler reflects red light and absorbs all other colours. Therefore the ruler appears to be red.

Step 3 : What pigments need to be mixed to get red?Red pigment is produced when magenta and yellow pigments are mixed.

## If we shine white light on a sheet of paper that can only reflect green light, what is the colour of the paper?

Since the colour of an object is determined by that frequency of light that is reflected, the sheet of paper will appear green, as this is the only frequency that is reflected. All the other frequencies are absorbed by the paper.

Since the colour of an object is determined by that frequency of light that is reflected, the sheet of paper will appear green, as this is the only frequency that is reflected. All the other frequencies are absorbed by the paper.

See less## The cover of a book appears to have a magenta colour. What colours of light does it reflect and what colours does it absorb?

We know that magenta is a combination of red and blue primary colours of light. Therefore the object must be reflecting blue and red light and absorb green.

We know that magenta is a combination of red and blue primary colours of light. Therefore the object must be reflecting blue and red light and absorb green.

See less## If cyan light shines on a dress that contains a pigment that is capable of absorbing blue, what colour does the dress appear?

Step 1 : Determine the component colours of cyan light Cyan light is made up of blue and green light. Step 2 : Determine solution If the dress absorbs the blue light then the green light must be reflected, so the dress will appear green!

See lessStep 1 : Determine the component colours of cyan lightCyan light is made up of blue and green light.

Step 2 : Determine solutionIf the dress absorbs the blue light then the green light must be reflected, so the dress will appear green!

## What colours of light are absorbed by a green pigment?

If the pigment is green, then green light must be reflected. Therefore, red and blue light are absorbed.

If the pigment is green, then green light must be reflected. Therefore, red and

See lessblue light are absorbed.

## If white light is shone through a glass plate that absorbs light of all frequencies except red, what is the colour of the glass plate?

Since the colour of an object is determined by that frequency of light that is transmitted, the glass plate will appear red, as this is the only frequency that is not absorbed.

Since the colour of an object is determined by that frequency of light that is transmitted, the glass plate will appear red, as this is the only frequency that is not absorbed.

See less## Calculate the de Broglie wavelength of an electron moving at 40 m·s −1.

Step 1 : Determine what is required and how to approach the problem We are required to calculate the de Broglie wavelength of an electron given its speed. We can do this by using: λ = h/mv Step 2 : Determine what is given We are given: • The velocity of the electron v = 40 m · s−1 and we know: • TheRead more

Step 1 : Determine what is required and how to approach the problemWe are required to calculate the de Broglie wavelength of an

electron given its speed. We can do this by using:

λ = h/mv

Step 2 : Determine what is givenWe are given:

• The velocity of the electron v = 40 m · s

^{−1}and we know:

• The mass of the electron me = 9,11 × 10

^{−31 }kg• Planck’s constant h = 6,63 × 10

^{−34 }J · sStep 3 : Calculate the de Broglie wavelengthλ = h/mv

=( 6,63 × 10

^{−34}J · s)/(9,11 × 10^{−31}kg)(40 m · s−1)= 1,82 × 10−5 m

= 0,0182 mm

Although the electron and cricket ball in the two previous examples are travelling at the same velocity the de Broglie wavelength of the electron is much larger than that of the cricket ball. This is because the wavelength is inversely proportional to the mass of the particle.

See less## A slit with a width of 2511 nm has green light of wavelength 532 nm impinge on it. The diffracted light interferes on a surface, at what angle will the first minimum be?

Step 1 : Check what you are given We know that we are dealing with interference patterns from the diffraction of light passing through a slit. The slit has a width of 2511 nm which is 2511 × 10−9 m and we know that the wavelength of the light is 532 nm which is 532 × 10−9 m. We are looking to determRead more

See lessStep 1 : Check what you are givenWe know that we are dealing with interference patterns from the

diffraction of light passing through a slit. The slit has a width of

2511 nm which is 2511 × 10−

^{9}m and we know that the wavelength of the light is 532 nm which is 532 × 10−^{9}m. We arelooking to determine the angle to first minimum so we know that

m = 1.

Step 2 : Applicable principlesWe know that there is a relationship between the slit width, wavelength and interference minimum angles:

sin θ = mλ

a

We can use this relationship to find the angle to the minimum by

substituting what we know and solving for the angle.

Step 3 : Substitutionsin θ = (532 × 10−

^{9}m)/(2511 × 10−^{9}m)sin θ = 532/2511

sin θ = 0.211867782

θ = sin−

^{1}0.211867782θ = 12.2°

The first minimum is at 12.2 degrees from the centre peak.

## A slit with a width of 2511 nm has red light of wavelength 650 nm impinge on it. The diffracted light interferes on a surface. At which angle will the first minimum be?

Step 1 : Check what you are given We know that we are dealing with interference patterns from the diffraction of light passing through a slit. The slit has a width of 2511 nm which is 2511 × 10−9 m and we know that the wavelength of the light is 650 nm which is 650 × 10−9 m. We are looking to determRead more

Step 1 : Check what you are givenWe know that we are dealing with interference patterns from the

diffraction of light passing through a slit. The slit has a width of

2511 nm which is 2511 × 10−

^{9}m and we know that the wavelength of the light is 650 nm which is 650 × 10−^{9}m. We arelooking to determine the angle to first minimum so we know that

m = 1.

See lessStep 2 : Applicable principlesWe know that there is a relationship between the slit width, wavelength and interference minimum angles:

sin θ = mλ

a

We can use this relationship to find the angle to the minimum by

substituting what we know and solving for the angle.

Step 3 : Substitutionsin θ = (650 × 10−

^{9}m)/(2511 × 10−^{9}m)sin θ = 650/2511

sin θ = 0.258861012

θ = sin−

^{1}0.258861012θ = 15°

The first minimum is at 15 degrees from the centre peak.