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Chapter 23: Problem 89

\(\bullet$$\bullet\) 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?

Short Answer

Expert verified
(a) 11.54掳; (b) 48.75掳.

Step by step solution

01

Understand the Concept

The problem requires understanding the concept of refraction and total internal reflection in optical mediums with different indices of refraction. We will use Snell's Law, which states that \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \), to find the critical angle where total internal reflection occurs.
02

Determine Critical Angle at Ice-Water Interface

At the ice-water interface, the light ray passes from water \((n_2 = 1.333)\) to ice \((n_1 = 1.309)\). We need to find the critical angle \( \theta_c \) using Snell's Law with \( \theta_2 = 90^\circ \) (angle in air). Setting \( \sin \theta_2 = 1 \), we solve \( n_2 \sin \theta_c = n_1 \) at the critical angle.
03

Calculate Critical Angle

Using the formula for the critical angle, \( \sin \theta_c = \frac{n_1}{n_2} = \frac{1.309}{1.333} \). Hence, \( \theta_c = \arcsin \left( \frac{1.309}{1.333} \right) \). Calculate this angle using a calculator.
04

Solve for Air-Water Interface Without Ice

After the ice melts, the light travels directly from water \((n = 1.333)\) to air \((n = 1.0)\). Use \( \sin \theta_c = \frac{n_{air}}{n_{water}} = \frac{1}{1.333} \) to find the new critical angle. Calculate \( \theta_c = \arcsin \left( \frac{1}{1.333} \right) \).
05

Conclude with Calculated Angles

With the values calculated: for ice-water interface, \( \theta_c \approx \arcsin(0.98) \) which is approximately 11.54掳. After ice melts, \( \theta_c \approx 48.75掳 \).

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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 fundamental concept in optics, especially when dealing with light passing through different mediums. It is defined as the angle of incidence beyond which light is completely reflected back into the medium, instead of passing through the interface. This phenomenon occurs when light travels from a medium with a higher index of refraction to one with a lower index.
  • For example, imagine light passing from water, with an index of refraction of 1.333, into air, having an index of 1.0. The critical angle is when the refraction turns into total internal reflection, and no light escapes into the air.
  • To calculate the critical angle, you can use Snell's Law. Set the angle of refraction to 90掳, meaning the refracted ray grazes along the interface. This simplifies Snell's Law to: \( n_1 \sin \theta_c = n_2 \sin 90^\circ \), giving the critical angle formula \( \sin \theta_c = \frac{n_2}{n_1} \).
Total Internal Reflection
Total internal reflection (TIR) is a fascinating occurrence in optics that happens when the light hits the boundary of two mediums at an angle greater than the critical angle. In this scenario, instead of refracting, the light is entirely reflected back into the original medium.
  • This can only occur when light moves from a medium with a higher index of refraction toward a medium with a lower one. It's commonly observed in optical fibers, where light travels by bouncing off the fiber's sides without escaping, thanks to TIR.
  • Take for instance the scenario in the exercise: the light traveling from water through ice before meeting the air. If the angle of incidence exceeds the calculated 11.54掳, light will undergo total internal reflection instead of moving into the air.
Refraction in Optics
Refraction is the bending of light as it passes from one transparent medium to another. This occurs due to a change in the light's speed, which depends on the indices of refraction of both media it is transitioning between.
  • When light enters a medium with a higher index of refraction (like going from air to water), it bends towards the normal line, slowing down in the process. Conversely, moving to a medium with a lower index causes the light to speed up and bend away from the normal.
  • This principle is central in the given problem as the light moves from water to ice, altering its path. It changes again when the ice is absent, and the light moves directly into air, recalculating the critical angles using Snell's Law.
Indices of Refraction
Indices of refraction (n) are values that indicate how much light slows down when passing through a material. The index of refraction is a key element in predicting how light bends when moving between different substances.
  • The higher the index, the more the light slows, and the more it bends. Water has an index of 1.333, ice is slightly less at 1.309, and air is 1.0, which is why their interfaces behave differently when light passes through.
  • The exercise shows how the indices affect the critical angle and the conditions for total internal reflection as light interacts with different materials. Particularly, calculating these angles helps predict whether light will pass through or reflect back depending on the medium's indices.

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Most popular questions from this chapter

\(\bullet$$\bullet\) A powerful searchlight shines on a man. The man's cross- sectional area is 0.500 \(\mathrm{m}^{2}\) perpendicular to the light beam, and the intensity of the light at his location is 36.0 \(\mathrm{kW} / \mathrm{m}^{2}\) . He is wearing black clothing, so that the light incident on him is totally absorbed. What is the magnitude of the force the light beam exerts on the man? Do you think he could sense this force?

\(\bullet\) The critical angle for total internal reflection at a liquid-air interface is \(42.5^{\circ} .\) (a) If a ray of light traveling in the liquid has an angle of incidence of \(35.0^{\circ}\) at the interface, what angle does the refracted ray in the air make with the normal? (b) If a ray of light traveling in air has an angle of incidence of \(35.0^{\circ}\) at the interface, what angle does the refracted ray in the liquid make with the normal?

\(\bullet$$\bullet\) Laser surgery. Very short pulses of high-intensity laser beams are used to repair detached portions of the retina of the eye. The brief pulses of energy absorbed by the retina welds the detached portion back into place. In one such procedure, a laser beam has a wavelength of 810 \(\mathrm{nm}\) and delivers 250 \(\mathrm{mW}\) of power spread over a circular spot 510\(\mu \mathrm{m}\) in diameter. The vitreous humor (the transparent fluid that fills most of the eye) has an index of refraction of 1.34 . (a) If the laser pulses are each 1.50 \(\mathrm{ms}\) long, how much energy is delivered to the retina with each pulse? (b) What average pressure does the pulse of the laser beam exert on the retina as it is fully absorbed by the circular spot? (c) What are the wavelength and frequency of the laser light inside the vitreous humor of the eye? (d) What are the maximum values of the electric and magnetic fields in the laser beam?

\(\bullet\) Unpolarized light with intensity \(I_{0}\) is incident on an ideal polarizing filter. The emerging light strikes a second ideal polarizing filter whose axis is at \(41.0^{\circ}\) to that of the first. Determine (a) the intensity of the beam after it has passed through the second polarizer and (b) its state of polarization.

\(\bullet\) Threshold of vision. Under controlled darkened conditions in the laboratory, a light receptor cell on the retina of a person's eye can detect a single photon (more on photons in Chapter 28 ) of light of wavelength 505 \(\mathrm{nm}\) and having an energy of \(3.94 \times 10^{-19} \mathrm{J} .\) We shall assume that this energy is absorbed by a single cell during one period of the wave. Cells of this kind are called rods and have a diameter of approximately 0.0020 \(\mathrm{mm} .\) What is the intensity (in \(\mathrm{W} / \mathrm{m}^{2} )\) delivered to a rod?
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