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A ray of light is traveling in a glass cube that is totally immersed in water. You find that if the ray is incident on the glass-water interface at an angle to the normal larger than \(48.7^{\circ}\) , no light is refracted into the water. What is the refractive index of the glass?

Short Answer

Expert verified
The refractive index of the glass is approximately 1.775.

Step by step solution

01

Understanding the Critical Angle

The given problem states that no light is refracted into the water at angles larger than \(48.7^{\circ}\). This means that \(48.7^{\circ}\) is the critical angle where light experiences total internal reflection. We need to calculate the refractive index of the glass using the known refractive index of water and this critical angle.
02

Utilize Snell's Law

To find the refractive index of the glass, apply Snell's Law at the critical angle: \( n_{glass} \sin(\theta_c) = n_{water} \sin(90^{\circ}) \). Here, \(n_{water} = 1.33\) represents the refractive index of water and \(\theta_c = 48.7^{\circ}\) is the critical angle.
03

Substitute Values and Solve

Substitute the values into Snell's Law: \( n_{glass} \sin(48.7^{\circ}) = 1.33 \sin(90^{\circ}) \). Simplify this to get \( n_{glass} \sin(48.7^{\circ}) = 1.33 \), and rearrange it to solve for \( n_{glass} \):\[ n_{glass} = \frac{1.33}{\sin(48.7^{\circ})}\].
04

Calculate the Refractive Index of Glass

Calculate \(\sin(48.7^{\circ})\), which is approximately \(0.749\). Substitute this value back into the equation:\[ n_{glass} = \frac{1.33}{0.749} \approx 1.775 \].
05

Conclusion

The refractive index of the glass is calculated using the given critical angle and the refractive index of water. With the steps outlined, we have found that the refractive index of the glass is approximately 1.775.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Snell's Law
When light travels between two different media, its speed and direction change. This process is known as refraction. Snell's Law helps us understand how light bends when crossing from one medium to another. The law is mathematically expressed by the equation:
  • \(n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\)
Here, \(n_1\) and \(n_2\) are the refractive indices of the respective media, while \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction measured from the normal. Refractive indices are unique to each medium, like glass or water.

Understanding Snell’s Law is crucial for calculating the behavior of light at the interface of materials. In our exercise, we apply Snell's Law to determine the critical angle—a concept that depends on these indices. When the light crosses the boundary at this specific angle, it results in total internal reflection instead of refraction.
Critical Angle
The critical angle is the specific angle of incidence at which light travels along the boundary of two different media and no refraction occurs. Instead, light is perfectly refracted back into the original medium, a phenomenon known as the critical angle.
  • The critical angle only exists when light travels from a medium with a higher refractive index to one with a lower refractive index. In our scenario, the light moves from glass to water, where glass has a higher refractive index than water.
When the incidence angle exceeds this critical angle, as noted in our exercise summary where the angle is cited as \(48.7^{\circ}\), light cannot refract into the second medium, and total internal reflection occurs.

Calculating the critical angle helps engineers and scientists in designing fiber optic cables and other technologies that rely on light propagation.
Total Internal Reflection
Total internal reflection is a phenomenon occurring when a light wave travels from a medium with a higher refractive index to one with a lower refractive index at an angle greater than the critical angle. In this process, no light refracts out of the original medium. Instead, it reflects back into the medium completely.
  • This is a principle used in fiber optics, where light signals are transmitted over long distances with minimal loss.
  • It also explains why a straw looks bent when partially submerged in water due to the way light travels differently in water and air.
In the context of the exercise, the fact that no light refracted into the water from the glass at angles greater than \(48.7^{\circ}\) indicates total internal reflection. This phenomenon allows for efficient light transmission, which finds applications in various optical devices.
Glass and Water Interface
The interaction at the glass and water interface is an excellent example for exploring refractive phenomena like total internal reflection and critical angle. At this interface, light exhibits unique behaviors due to the different refractive indices.
  • Glass, typical in many optical settings, generally has a higher refractive index than water.
  • This makes it a great medium to observe total internal reflection.
In the provided exercise, the glass and water interface demonstrate these principles as light transitions from a denser medium to a less dense one. Calculating behaviors like the critical angle helps predict how light will travel at this boundary. Mastering these concepts lets students and professionals manipulate light in various scientific, technical, and engineering contexts for innovative applications such as lenses, watercraft sensors, and photographic equipment.

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Most popular questions from this chapter

A beaker with a mirrored bottom is filled with a liquid whose index of refraction is \(1.63 .\) A light beam strikes the top surface of the liquid at an angle of \(42.5^{\circ}\) from the normal. At what angle from the normal will the beam exit from the liquid after traveling down through the liquid, reflecting from the mirrored bottom, and returning to the surface?

A thin layer of ice \((n=1.309)\) floats on the surface of water \((n=1.333)\) in a bucket. A ray of light from the bottom of the bucket travels upward through the water. (a) What is the largest angle with respect to the normal that the ray can make at the ice-water interface and still pass out into the air above the ice? (b) What is this angle after the ice melts?

Old photographic plates were made of glass with a light-sensitive emulsion on the front surface. This emulsion was some what transparent. When a bright point source is focused on the front of the plate, the developed photograph will show a halo around the image of the spot. If the glass plate is 3.10 \(\mathrm{mm}\) thick and the halos have an inner radius of \(5.34 \mathrm{mm},\) what is the index of refraction of the glass? (Hint: Light from the spot on the front surface is scattered in all directions by the emulsion. Some of it is then totally reflected at the back surface of the plate and returns to the front surface.)

Light with a frequency of \(5.80 \times 10^{14} \mathrm{Hz}\) travels in a block of glass that has an index of refraction of \(1.52 .\) What is the wave-length of the light (a) in vacuum and (b) in the glass?

A ray of light is incident on a plane surface separating two sheets of glass with refractive indexes 1.70 and 1.58 . The angle of incidence is \(62.0^{\circ}\) , and the ray originates in the glass with \(n=1.70 .\) Compute the angle of refraction.

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