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Chapter 33: Problem 16

A flat piece of glass covers the top of a vertical cylinder that is completely filled with water. If a ray of light traveling in the glass is incident on the interface with the water at an angle of \(\theta_a = 36.2{^\circ}\), the ray refracted into the water makes an angle of 49.8\(^\circ\) with the normal to the interface. What is the smallest value of the incident angle \(\theta_a\) for which none of the ray refracts into the water?

Short Answer

Expert verified
The smallest incident angle for no refraction is 62.5掳.

Step by step solution

01

Understand Snell's Law

Snell's Law relates the angles of incidence and refraction to the refractive indices of the two media. It is given by \( n_{1} \sin(\theta_{1}) = n_{2} \sin(\theta_{2}) \), where \( n_{1} \) and \( n_{2} \) are the refractive indices of glass and water respectively, and \( \theta_{1} \) and \( \theta_{2} \) are the angles of incidence and refraction.
02

Identify Given Values

From the problem, \( \theta_{a} = 36.2{^\circ} \) is the angle of incidence, and \( \theta_{2} = 49.8{^\circ} \) is the angle of refraction. \( n_{1} = 1.50 \) for glass and \( n_{2} = 1.33 \) for water. Use these values to replace \( \theta_{1} \) and other variables appropriately in Snell's Law.
03

Calculate Critical Angle for Total Internal Reflection

The critical angle \( \theta_{c} \) is the angle of incidence above which light cannot refract into the other medium, given by \( \theta_{c} = \arcsin \left( \frac{n_{2}}{n_{1}} \right) \). Calculate \( \theta_{c} = \arcsin \left( \frac{1.33}{1.50} \right) \).
04

Compute Result

Using a calculator, find \( \theta_{c} = \arcsin \left( \frac{1.33}{1.50} \right) \approx 62.5{^\circ} \). Therefore, if the incident angle \( \theta_{a} \) is greater than \( 62.5{^\circ} \), none of the light will refract into the water.

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Key Concepts

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

Critical Angle
The critical angle is a concept linked to the phenomenon of total internal reflection. It is defined as the smallest angle of incidence in the denser medium (like glass) for which the light refracts along the boundary into the less dense medium (like water). Beyond this angle, light no longer exits into the less dense medium but is entirely reflected back.

To find the critical angle, we use Snell's Law, which is given by:\[ n_{1} \sin(\theta_{1}) = n_{2} \sin(\theta_{2})\]For the critical angle \(\theta_{c}\), \(\theta_{2}\) becomes \(90{^\circ}\), making \(\sin(\theta_{2}) = 1\). Thus,\[ \theta_{c} = \arcsin\left( \frac{n_{2}}{n_{1}} \right)\]This formula allows us to calculate the critical angle between any two media for total internal reflection to occur.
Total Internal Reflection
Total internal reflection occurs when a light wave traveling through a denser medium hits the boundary with a less dense medium at an angle greater than the critical angle. Instead of refracting into the second medium, the light is reflected entirely back into the first medium.

This phenomenon is used in various applications:
  • Fiber optic cables, where light signals reflect internally to transmit data over long distances.
  • Binoculars and periscopes, which rely on total internal reflection to redirect light paths.
Total internal reflection not only makes these technologies possible but also ensures they work efficiently and effectively.
Refractive Index
The refractive index is a measure of how much the speed of light is reduced inside a medium compared to its speed in a vacuum. It is a dimensionless number, often denoted by \(n\). The refractive index determines how much the path of light bends, or refracts, when entering a medium.

Each material has its own refractive index:
  • Glass has a refractive index of about 1.50.
  • Water has a refractive index around 1.33.
To calculate how light will bend when passing from one medium to another, we use Snell's Law:\[ n_{1} \sin(\theta_{1}) = n_{2} \sin(\theta_{2})\]Understanding refractive indices is crucial in designing optical systems and understanding how light behaves at interfaces.

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Most popular questions from this chapter

Light with a frequency of \(5.80 \times 10^{14}\) Hz travels in a block of glass that has an index of refraction of 1.52. What is the wavelength 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.

When the sun is either rising or setting and appears to be just on the horizon, it is in fact below the horizon. The explanation for this seeming paradox is that light from the sun bends slightly when entering the earth鈥檚 atmosphere, as shown in Fig. \(\textbf{P33.51}\). Since our perception is based on the idea that light travels in straight lines, we perceive the light to be coming from an apparent position that is an angle \(\delta\) above the sun's true position. (a) Make the simplifying assumptions that the atmosphere has uniform density, and hence uniform index of refraction \(n\), and extends to a height \(h\) above the earth's surface, at which point it abruptly stops. Show that the angle \(\delta\) is given by $${ \delta = \mathrm{arcsin} ({{nR}\over{R + h}}}) - \mathrm{arcsin}({{{R}\over R + h}}) $$ where \(R\) = 6378 km is the radius of the earth. (b) Calculate \(\delta\) using \(n\) = 1.0003 and \(h\) = 20 km. How does this compare to the angular radius of the sun, which is about one quarter of a degree? (In actuality a light ray from the sun bends gradually, not abruptly, since the density and refractive index of the atmosphere change gradually with altitude.)

A light beam travels at \(1.94 \times 10^8\) m/s in quartz. The wavelength of the light in quartz is 355 nm. (a) What is the index of refraction of quartz at this wavelength? (b) If this same light travels through air, what is its wavelength there?

The refractive index of a certain glass is 1.66. For what incident angle is light reflected from the surface of this glass completely polarized if the glass is immersed in (a) air and (b) water?
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