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The drawing shows a rectangular block of glass \((n=1.52)\) surrounded by liquid carbon disulfide \((n=1.63) .\) A ray of light is incident on the glass at point A with a \(30.0^{\circ}\) angle of incidence. At what angle of refraction does the ray leave the glass at point \(\mathrm{B} ?\)

Short Answer

Expert verified
The ray leaves the glass at an angle of refraction of approximately \(25.8^{\circ}\).

Step by step solution

01

Understand the Context

We have a rectangular block of glass with an index of refraction \( n_1 = 1.52 \) and it is surrounded by liquid carbon disulfide with an index of refraction \( n_2 = 1.63 \). A light ray strikes the glass surface at point A with an angle of incidence of \(30.0^{\circ}\). We need to find the angle of refraction as the light exits the glass at point B.
02

Apply Snell's Law at Point A

Use Snell's Law, \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \), at point A where \( \theta_1 = 30.0^{\circ} \): \[ 1.52 \sin(30.0^{\circ}) = 1.63 \sin(\theta) \]Calculate \( \sin(30.0^{\circ}) \), which is \(0.5\), and solve for \( \theta \): \[ 1.52 \times 0.5 = 1.63 \sin(\theta) \] \[ 0.76 = 1.63 \sin(\theta) \] \[ \sin(\theta) = \frac{0.76}{1.63} \approx 0.4663 \] Then compute the angle \( \theta \) which is \(\arcsin(0.4663) \approx 27.8^{\circ}\).
03

Light Travels Through the Glass

Inside the glass, the angle of refraction calculated from point A becomes the angle of incidence at the surface where the glass meets the carbon disulfide (point B). Therefore, \( \theta_B = 27.8^{\circ} \).
04

Apply Snell's Law at Point B

Again apply Snell's Law at point B where the light exits the glass and enters the liquid carbon disulfide:\[ n_1 \sin(\theta_B) = n_2 \sin(\theta_r) \] Substitute \( \theta_B = 27.8^{\circ}\) into the equation: \[ 1.52 \sin(27.8^{\circ}) = 1.63 \sin(\theta_r) \] \[ 1.52 \times 0.4676 = 1.63 \sin(\theta_r) \] \[ 0.7109 = 1.63 \sin(\theta_r) \] \[ \sin(\theta_r) = \frac{0.7109}{1.63} \approx 0.436 \] Calculate \( \theta_r = \arcsin(0.436) \approx 25.8^{\circ} \).
05

Conclusion

The angle of refraction \( \theta_r \) as the ray exits the glass at point B into the surrounding carbon disulfide is approximately \( 25.8^{\circ} \).

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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 a fundamental principle in understanding how light behaves. It describes how light bends, or refracts, as it passes from one medium into another. The formula for Snell's Law is given by: \[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \] This law relates the indices of refraction \( n_1 \) and \( n_2 \) of two different media, and the angles of incidence \( \theta_1 \) and refraction \( \theta_2 \).
  • Medium 1 is where the light originates and Medium 2 is the new medium it enters.
  • \( n_1 \) and \( n_2 \) indicate how much light slows down in each medium.
Snell's Law is crucial for calculating how much the light path will shift. This happens because materials like water, glass, or air have different optical densities. The light bends towards the normal if it enters a denser medium, meaning a higher index of refraction. Conversely, it bends away from the normal if it moves into a less dense medium.
Index of Refraction
The index of refraction, often denoted as \( n \), shows how fast light travels in a material. It is a dimensionless number representing the ratio of the speed of light in a vacuum to the speed of light in the material itself. Mathematically, this is represented as: \[ n = \frac{c}{v} \] where \( c \) is the speed of light in a vacuum, and \( v \) is the speed of light in the material.
  • An index of refraction greater than 1 indicates that light travels slower in the material than in a vacuum.
  • Higher values suggest a material is optically denser.
In our exercise, glass has an index of refraction \( n = 1.52 \), and liquid carbon disulfide \( n = 1.63 \). Light travels slower in these materials than in a vacuum. Thus, when light moves between them, refraction occurs. Knowing the values of \( n \) is essential for using Snell's Law effectively.
Angle of Incidence
The angle of incidence is the angle between the incident light ray hitting a surface and the line perpendicular to that surface, called the normal line. This angle is crucial for understanding refraction.
  • It is always measured from the normal, not the surface itself.
  • The angle of incidence helps determine how much light will bend upon moving from one medium to another.
In the given exercise, the angle of incidence is \( 30.0^{\circ} \) as the light enters the glass from the surrounding carbon disulfide. The geometry of this setup influences how the light ray bends as it travels through various substances. By measuring from the normal, predictions about the behavior of light using Snell's Law can be made.
Angle of Refraction
The angle of refraction is the angle between the refracted light ray and the normal line in the new medium. This angle represents how much the light path has changed from its original direction.
  • It is determined using Snell's Law once the angle of incidence and the indices of refraction are known.
  • The angle of refraction allows calculating the light's direction as it exits or moves through materials.
In our exercise, light exits the glass into the carbon disulfide at an angle of refraction of \( 25.8^{\circ} \). This was found by applying Snell's Law at both interfaces (from carbon disulfide to glass, and then glass to carbon disulfide). Correctly finding this angle is essential for understanding the path light takes when crossing different mediums.

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Most popular questions from this chapter

A ray of sunlight is passing from diamond into crown glass; the angle of incidence is \(35.00^{\circ} .\) The indices of refraction for the blue and red components of the ray are: blue \(\left(n_{\text {diamond }}=2.444, n_{\text {crown glass }}=1.531\right),\) and red \(\left(n_{\text {diamond }}=2.410, n_{\text {crown glass}}=1.520\right) .\) Determine the angle between the refracted blue and red rays in the crown glass.

Two converging lenses are separated by \(24.00 \mathrm{cm} .\) The focal length of each lens is \(12.00 \mathrm{cm} .\) An object is placed \(36.00 \mathrm{cm}\) to the left of the lens that is on the left. Determine the final image distance relative to the lens on the right.

A camera is supplied with two interchangeable lenses, whose focal lengths are 35.0 and \(150.0 \mathrm{mm} .\) A woman whose height is \(1.60 \mathrm{m}\) stands \(9.00 \mathrm{m}\) in front of the camera. What is the height (including sign) of her image on the image sensor, as produced by (a) the 35.0 -mm lens and (b) the 150.0 -mm lens?

An object is placed \(20.0 \mathrm{cm}\) to the left of a diverging lens \((f=\) \(-8.00 \mathrm{cm}$$) .\) A concave mirror \((f=12.0 \mathrm{cm})\) is placed \(30.0 \mathrm{cm}\) to the right of the lens. (a) Find the final image distance, measured relative to the mirror. (b) Is the final image real or virtual? (c) Is the final image upright or inverted with respect to the original object?

A converging lens has a focal length of \(88.00 \mathrm{cm}\). An object \(13.0 \mathrm{cm}\) tall is located \(155.0 \mathrm{cm}\) in front of this lens. (a) What is the image distance? (b) Is the image real or virtual? (c) What is the image height? Be sure to include the proper algebraic sign.

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