A beam of light of two different wavelengths enters a pane of glass 3.00 mm thick at an angle of incidence of 56°. The indices of refraction for the two different colors are 1.512 and 1.524.
a) Suppose one is red light and the other is blue light, which one will be refracted more (or bend more)?
b) Because of dispersion, the colored beams, are separated by a small distance. How far apart are they in distance when they exit the plane of glass?
The Correct Answer and Explanation is:
To answer the questions regarding the refraction of light through a pane of glass, we will first understand how light behaves when passing from one medium to another, particularly with respect to the refractive index.
Part A: Which Color Bends More?
When light passes from air into glass, it slows down and changes direction due to refraction. The amount by which light bends depends on the angle of incidence and the indices of refraction of the two media. According to Snell’s Law:
[
n_1 \sin(\theta_1) = n_2 \sin(\theta_2)
]
where:
- ( n_1 ) is the refractive index of the first medium (air, approximately 1.00),
- ( \theta_1 ) is the angle of incidence,
- ( n_2 ) is the refractive index of the second medium (glass),
- ( \theta_2 ) is the angle of refraction.
Given the indices of refraction for red light (( n_{red} = 1.512 )) and blue light (( n_{blue} = 1.524 )), we can conclude that the blue light, having a higher refractive index, will bend more than red light. This is because a higher index of refraction indicates a greater change in speed and direction of light as it enters the glass. Therefore, the answer to Part A is blue light bends more.
Part B: Distance Apart When Exiting the Glass
To find out how far apart the two colors of light are when they exit the pane of glass, we can use the thickness of the glass and the angles of refraction.
- Calculate the angles of refraction for both colors using Snell’s Law: For red light:
[
n_{air} \sin(56^\circ) = n_{red} \sin(\theta_{red})
]
[
\sin(\theta_{red}) = \frac{n_{air} \sin(56^\circ)}{n_{red}} \approx \frac{1.00 \times 0.829}{1.512} \approx 0.548 \Rightarrow \theta_{red} \approx 33.4^\circ
] For blue light:
[
n_{air} \sin(56^\circ) = n_{blue} \sin(\theta_{blue})
]
[
\sin(\theta_{blue}) = \frac{n_{air} \sin(56^\circ)}{n_{blue}} \approx \frac{1.00 \times 0.829}{1.524} \approx 0.544 \Rightarrow \theta_{blue} \approx 32.8^\circ
] - Calculate the distance traveled within the glass. The path length of the light inside the glass can be determined using the thickness ( d = 3.00 \, \text{mm} ) and the angles of refraction. The horizontal distance ( x ) covered by each color as it exits can be calculated by: [
x = d \tan(\theta)
] For red light:
[
x_{red} = 3.00 \, \text{mm} \times \tan(33.4^\circ) \approx 3.00 \, \text{mm} \times 0.649 \approx 1.95 \, \text{mm}
] For blue light:
[
x_{blue} = 3.00 \, \text{mm} \times \tan(32.8^\circ) \approx 3.00 \, \text{mm} \times 0.634 \approx 1.90 \, \text{mm}
] - Determine the distance apart when exiting:
[
\Delta x = x_{red} – x_{blue} \approx 1.95 \, \text{mm} – 1.90 \, \text{mm} = 0.05 \, \text{mm}
]
Thus, the two beams of light will be separated by approximately 0.05 mm when they exit the pane of glass. This small separation illustrates the effect of dispersion, where different wavelengths of light refract at slightly different angles due to their differing indices of refraction, resulting in the separation of colors.