Phy 124 – Chapter 22 Problems
1. Consider the following.
A spot on a boat is 2.5 m above the water, and the light strikes the water at a point that is 8.0 m horizontally displaced from the spotlight. The depth of the water is 4.0 m. Determine the distance d, which locates the point where the light strikes the bottom. (The index of refraction for water is 1.333)
2. A source emits monochromatic light of wavelength 495nm in air. When light passes through a liquid, its wave length reduces to434nm. What is the liquids index of refraction?
3. a) Derive an expression for the wavelength λ of a photon in terms of its energy E, Planck’s constant h and the speed of light c.
b) What does the equation say about the wavelength of higher-energy photons?
To solve these problems, we will need to use relevant formulas and physics principles.
1. The problem involves the refraction of light at the interface between air and water. We need to find the distance d, which locates the point where the light strikes the bottom.
To solve this, we can use Snell's Law, which relates the angles of incidence and refraction and the indices of refraction of the two media:
n1 * sin(theta1) = n2 * sin(theta2)
where n1 and n2 are the indices of refraction of the initial and final media, and theta1 and theta2 are the angles of incidence and refraction, respectively.
In this case, our initial medium is air and the final medium is water, so n1 = 1 (index of refraction of air) and n2 = 1.333 (index of refraction of water).
Looking at the problem, the incident light is traveling horizontally, so we have theta1 = 0 degrees (or sin(theta1) = 0). The angle of refraction theta2 will depend on the path the light takes in the water.
We can use the fact that the depth of the water is 4.0 m and the spot on the boat is 2.5 m above the water. The horizontal displacement of the light is given as 8.0 m.
Using trigonometry, we can determine the angle of refraction theta2:
sin(theta2) = (d - 2.5 m) / 4.0 m
Now, plugging the values into Snell's Law:
1 * 0 = 1.333 * sin(theta2)
Simplifying the equation:
sin(theta2) = 0
This implies that the angle of refraction is 0 degrees since the sin(0) = 0.
Therefore, the light travels straight down without bending. The distance d, which locates the point where the light strikes the bottom, will be the horizontal displacement given by the problem, which is 8.0 m.
So, the distance d is 8.0 m.
2. This problem involves the change in wavelength of light as it passes through a liquid.
To determine the liquid's index of refraction, we can use the formula:
n = c/λ
where n is the index of refraction, c is the speed of light in vacuum, and λ is the wavelength.
Initially, the light has a wavelength of 495 nm in air. So, λ1 = 495 nm = 495 × 10^(-9) m.
After passing through the liquid, the wavelength reduces to 434 nm. So, λ2 = 434 nm = 434 × 10^(-9) m.
Now, using the formula for index of refraction:
n = c/λ
we can find the index of refraction of the liquid:
n = c/λ2 = (3 × 10^8 m/s)/(434 × 10^(-9) m)
Simplifying the equation:
n = 6.9120 ≈ 6.91
Therefore, the liquid's index of refraction is approximately 6.91.
3. a) This problem involves deriving an expression for the wavelength of a photon in terms of its energy, Planck's constant, and the speed of light.
The energy of a photon is given by the equation:
E = hc/λ
where E is the energy, h is Planck's constant (6.626 × 10^(-34) J·s), c is the speed of light (approximately 3 × 10^8 m/s), and λ is the wavelength.
We can rearrange this equation to solve for λ:
λ = hc/E
Therefore, the expression for the wavelength λ of a photon in terms of its energy E, Planck's constant h, and the speed of light c is λ = hc/E.
b) The equation λ = hc/E shows an inverse relationship between the wavelength (λ) and the energy (E) of a photon. As the energy of a photon increases, its wavelength decreases. This indicates that higher-energy photons have shorter wavelengths.