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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) Approximately 50.4°; (b) Approximately 48.8° after the ice melts.

Step by step solution

01

Understand the Concept

The problem asks for the largest angle at which a ray of light can pass from one medium to another without being totally internally reflected. This is known as the critical angle, which occurs when light is at the boundary between refraction and reflection.
02

Determine the Interface

Identify the interfaces involved: the light travels from water (\(n = 1.333\)) to ice (\(n = 1.309\)), and then from ice to air (\(n = 1.00\)). Calculate the critical angle for both interfaces.
03

Use Snell's Law

Snell's Law is given by \(n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\). To find the critical angle, we set \(\theta_2 = 90^\circ\) because beyond this angle, total internal reflection occurs. Hence, the equation becomes \(\sin(\theta_c) = \frac{n_2}{n_1}\).
04

Calculate Critical Angle at Ice-Water Interface Before Melting

For the ice-water interface, the ray of light is going from water to ice. The critical angle (\(\theta_{c1}\)) can be calculated using \(\sin(\theta_{c1}) = \frac{n_{ice}}{n_{water}} = \frac{1.309}{1.333}\). Calculate \(\theta_{c1}\) using this ratio.
05

Calculate Critical Angle at Ice-Air Interface

For the ice-air interface, the ray goes from ice to air. The critical angle (\(\theta_{c2}\)) is found using \(\sin(\theta_{c2}) = \frac{n_{air}}{n_{ice}} = \frac{1.00}{1.309}\). Calculate \(\theta_{c2}\) using this ratio.
06

Determine the Largest Angle Before Ice Melts

The angle that limits light exiting from water through ice into air is determined by the smaller critical angle from the steps above. Compare the angles \(\theta_{c1}\) and \(\theta_{c2}\) and choose the smaller as the largest angle of incidence for light passing out into the air.
07

Angle After Ice Melts

Once the ice melts, light directly travels from water to air. Calculate the critical angle using \(\sin(\theta_{c}) = \frac{n_{air}}{n_{water}} = \frac{1.00}{1.333}\). This gives the new maximum angle.

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Key Concepts

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

Snell's Law
Snell's Law is fundamental in understanding the refraction of light when it moves between different media. It explains why light bends when crossing from one substance to another. The law is 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 two media, and \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction, respectively.

One practical application of Snell's Law is in calculating the critical angle for total internal reflection, a crucial concept in optics. For a ray traveling from a denser medium to a less dense one, if it hits the boundary at an angle greater than this critical angle, it will not pass through but instead reflect back into the denser medium.
  • The denser medium has a higher refractive index (\(n\)).
  • Total internal reflection occurs when \(\theta_2\) is 90 degrees, making the refracted ray skim along the boundary.
Total Internal Reflection
Total Internal Reflection is an intriguing phenomenon where light is completely reflected back into a medium rather than passing through it. This effect is observed when a ray of light moves from a medium with a higher refractive index to one with a lower refractive index.

The critical angle is key to understanding total internal reflection. It is the maximum angle of incidence in the denser medium beyond which the light cannot refract into the less dense medium. Instead, it is entirely reflected back inside. This happening is not just a peculiarity; it is widely used in designing instruments like fiber optics, binoculars, and even periscopes.
  • The critical angle depends on the refractive indices of the two media in contact.
  • If the incident angle is greater than the critical angle, the light doesn't escape.
Refraction of Light
Refraction is the bending of light as it passes from one medium to another with a different refractive index. This occurs because the speed of light changes in different materials, altering its path.

When light enters a denser medium (e.g., from air to water), it slows down and bends towards the normal – an imaginary line perpendicular to the surface at the point of incidence. Conversely, transitioning into a less dense medium causes the ray to speed up and bend away from the normal. Snell's Law helps calculate this bending precisely using the media's refractive indices.
  • Bending is more pronounced with larger differences in refractive indices.
  • Light waves are generally fastest in vacuum and slowest in solid materials.

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Most popular questions from this chapter

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?

Physicians use high-frequency (\(f\) = 1\(-\)5 MHz) sound waves, called ultrasound, to image internal organs. The speed of these ultrasound waves is 1480 m\(/\)s in muscle and 344 m\(/\)s in air. We define the index of refraction of a material for sound waves to be the ratio of the speed of sound in air to the speed of sound in the material. Snell's law then applies to the refraction of sound waves. (a) At what angle from the normal does an ultrasound beam enter the heart if it leaves the lungs at an angle of 9.73\(^\circ\) from the normal to the heart wall? (Assume that the speed of sound in the lungs is 344 m\(/\)s.) (b) What is the critical angle for sound waves in air incident on muscle?

A parallel beam of unpolarized light in air is incident at an angle of 54.5\(^\circ\) (with respect to the normal) on a plane glass surface. The reflected beam is completely linearly polarized. (a) What is the refractive index of the glass? (b) What is the angle of refraction of the transmitted beam?

Optical fibers are constructed with a cylindrical core surrounded by a sheath of cladding material. Common materials used are pure silica \(n_2 = 1.4502\) for the cladding and silica doped with germanium \(n_1 = 1.4652\) for the core. (a) What is the critical angle \(\theta_{crit}\) for light traveling in the core and reflecting at the interface with the cladding material? (b) The numerical aperture (NA) is defined as the angle of incidence \(theta_i\) at the flat end of the cable for which light is incident on the core-cladding interface at angle \(\theta_{crit}\) (\(\textbf{Fig. P33.46}\)). Show that sin \(\theta_i\) =\( \sqrt {n^2_1 - n^2_2}\) . (c) What is the value of \(\theta_i\) for \(n_1\) = 1.465 and \(n_2\) = 1.450?

A parallel beam of light in air makes an angle of 47.5\(^\circ\) with the surface of a glass plate having a refractive index of 1.66. (a) What is the angle between the reflected part of the beam and the surface of the glass? (b) What is the angle between the refracted beam and the surface of the glass?

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