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Chapter 26: Problem 107

mmh A glass block \((n=1.56)\) is immersed in a liquid. A ray of light within the glass hits a glass-liquid surface at a \(75.0^{\circ}\) angle of incidence. Some of the light enters the liquid. What is the smallest possible refractive index for the liquid?

Short Answer

Expert verified
To find the minimum refractive index for the liquid, use Snell鈥檚 Law and the critical angle condition.

Step by step solution

01

Recognize Snell's Law

To solve this problem, we need to use Snell's Law, which relates the angles and the refractive indices of two media. Snell's Law is given by: \[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \]where \( n_1 \) and \( \theta_1 \) are the refractive index and angle of incidence in the first medium (glass in this case), and \( n_2 \) and \( \theta_2 \) are the refractive index and angle of refraction in the second medium (liquid).
02

Identify Critical Angle Condition

Since some light enters the liquid, it suggests that the angle of refraction is less than \(90^{\circ}\). For the light to pass into the second medium, the angle of incidence must be less than the critical angle, or exactly at the critical angle for minimal passing. The critical angle \( \theta_c \) occurs when the angle of refraction is \(90^{\circ}\), so we solve for \(n_2\) in: \[ \sin \theta_c = \frac{n_2}{n_1} \]

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Key Concepts

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

Refractive Index
The refractive index is a measure of how much a medium slows down light compared to air or vacuum. It is denoted by the symbol \( n \). Different materials have different refractive indices. For instance, glass commonly has a refractive index of around 1.5 or more, whereas air's refractive index is approximately 1.
  • A higher refractive index means that light travels slower in that medium.
  • The refractive index is unitless and is defined as the ratio between the speed of light in vacuum and the speed of light in the medium.
This concept is crucial in Snell's Law, which helps us understand the behavior of light as it moves between different materials.
Angle of Incidence
When a ray of light hits the surface between two different media, the angle it makes with the surface normal (an imaginary line perpendicular to the surface) is called the angle of incidence, denoted by \( \theta_1 \). It is critical in determining how much the light will bend when entering a new medium.
  • The angle of incidence is measured from the normal to the boundary.
  • In our exercise, the angle of incidence given for light passing from glass to liquid is \(75.0^{\circ}\).
A precise understanding of the angle of incidence helps predict the subsequent path of the light ray with accuracy.
Angle of Refraction
The angle of refraction, denoted as \( \theta_2 \), is the angle between the refracted ray and the normal. It occurs when light passes from one medium to another and changes direction, a phenomenon known as refraction.
  • The angle of refraction depends on the refractive indices of the two media and the angle of incidence.
  • According to Snell's Law, refracted light follows the equation \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \).
Snell's Law allows us to calculate the precise angle at which light refracts when transitioning into a new medium.
Critical Angle
The critical angle is a special angle of incidence in optics. It occurs when the angle of refraction reaches \(90^{\circ}\), making the refracted ray run along the boundary of the two media without entering the second material. Beyond the critical angle, light undergoes total internal reflection.
  • The critical angle depends on the refractive indices of the two materials involved.
  • It can be calculated using the relation \( \sin \theta_c = \frac{n_2}{n_1} \), where \( \theta_c \) is the critical angle, \( n_1 \) is the refractive index of the initial medium, and \( n_2 \) is that of the second medium.
Understanding the critical angle helps in applications like fiber optics, where light is trapped within a medium to transmit signals over long distances. In this exercise, calculating the critical angle is pivotal to determining the refractive index of the liquid for which the light still refracts.

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Most popular questions from this chapter

A telescope has an objective with a refractive power of 1.25 diopters and an eyepiece with a refractive power of 250 diopters. What is the angular magnification of the telescope?

The back wall of a home aquarium is a mirror that is a distance of 40.0 cm away from the front wall. The walls of the tank are negligibly thin. A fish, swimming midway between the front and back walls, is being viewed by a person looking through the front wall. The index of refraction of air is nair 1.000 and that of water is n water 1.333. (a) Calculate the apparent distance between the fish and the front wall. (b) Calculate the apparent distance between the image of the fish and the front wall. The index of refraction of air is \(n_{\mathrm{air}}=1.000\) and that of water is \(n_{\text { water }}=1.333 .\) (a) Calculate the apparent distance between the fish and the front wall. (b) Calculate the apparent distance between the image of the fish and the front wall.

A spectator, seated in the left-field stands, is watching a baseball player who is 1.9 m tall and is 75 m away. On a TV screen, located 3.0 m from a person watching the game at home, the image of this same player is 0.12 m tall. Find the angular size of the player as seen by (a) the spectator watching the game live and (b) the TV viewer. (c) To whom does the player appear to be larger?

A stamp collector is viewing a stamp with a magnifying glass held next to her eye. Her near point is 25 cm from her eye. (a) What is the refractive power of a magnifying glass that has an angular magnification of 6.0 when the image of the stamp is located at the near point? (b) What is the angular magnification when the image of the stamp is 45 cm from the eye?

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?
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