A beam of light of two different wavelengths enters a pane of glass 3.2 mm thick at an angle of incidence of 59°. The indices of refraction for the two different colors are 1.5 and 1.53. Because of dispersion, the colored beams, although parallel, are separated by a small distance. How far apart are they? (Give your answer to the nearest 0.0001 mm)
The Correct Answer and Explanation is :
The correct answer is: 0.0200 mm
To determine how far apart the two colored beams of light are after passing through a 3.2 mm thick pane of glass, we can use the principles of refraction and geometry. Given the indices of refraction for the two colors, ( n_1 = 1.5 ) and ( n_2 = 1.53 ), we can find the angles of refraction using Snell’s Law, which states:
[
n_1 \sin(\theta_1) = n_2 \sin(\theta_2)
]
Where:
- ( n_1 ) and ( n_2 ) are the indices of refraction of the first medium (air, approximately ( n \approx 1 )) and glass, respectively.
- ( \theta_1 ) is the angle of incidence (59°).
- ( \theta_2 ) is the angle of refraction.
- Calculate the angle of refraction for both wavelengths:
Using Snell’s Law for both colors:
[
\sin(\theta_1) = \sin(59°) \approx 0.5150
]
For the first wavelength (( n_1 = 1.5 )):
[
1 \cdot 0.5150 = 1.5 \sin(\theta_{2,1})
]
[
\sin(\theta_{2,1}) = \frac{0.5150}{1.5} \approx 0.3433
]
[
\theta_{2,1} \approx \arcsin(0.3433) \approx 20.1°
]
For the second wavelength (( n_2 = 1.53 )):
[
1 \cdot 0.5150 = 1.53 \sin(\theta_{2,2})
]
[
\sin(\theta_{2,2}) = \frac{0.5150}{1.53} \approx 0.3366
]
[
\theta_{2,2} \approx \arcsin(0.3366) \approx 19.6°
]
- Calculate the distances traveled through the glass:
Using the thickness of the glass (3.2 mm) and the angles of refraction, we can find the horizontal distance each beam travels within the glass:
For the first color:
[
d_1 = \frac{3.2 \, \text{mm}}{\cos(\theta_{2,1})} \cdot \sin(\theta_{2,1})
]
[
d_1 = \frac{3.2 \, \text{mm}}{\cos(20.1°)} \approx \frac{3.2 \, \text{mm}}{0.9397} \approx 3.41 \, \text{mm}
]
For the second color:
[
d_2 = \frac{3.2 \, \text{mm}}{\cos(\theta_{2,2})} \cdot \sin(\theta_{2,2})
]
[
d_2 = \frac{3.2 \, \text{mm}}{\cos(19.6°)} \approx \frac{3.2 \, \text{mm}}{0.9430} \approx 3.39 \, \text{mm}
]
- Determine the distance between the two beams:
The difference in the horizontal distances can now be calculated:
[
\Delta d = d_1 – d_2 \approx 3.41 \, \text{mm} – 3.39 \, \text{mm} = 0.02 \, \text{mm}
]
Thus, the two colored beams will be separated by approximately 0.0200 mm after passing through the glass.
Conclusion
The final separation of the beams of light of different wavelengths after passing through a glass pane of 3.2 mm thick at an angle of incidence of 59° is approximately 0.0200 mm. This phenomenon is a result of dispersion, where different wavelengths of light bend at slightly different angles due to their respective indices of refraction. This calculation demonstrates the interaction of light with materials and the principles of optics.