/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 A layer of liquid B floats on li... [FREE SOLUTION] | 91Ó°ÊÓ

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A layer of liquid B floats on liquid A. A ray of light begins in liquid A and undergoes total internal reflection at the interface between the liquids when the angle of incidence exceeds \(36.5^{\circ} .\) When liquid \(B\) is replaced with liquid \(C,\) total internal reclection occurs for angles of incidence greater than \(47.0^{\circ} .\) Find the ratio \(n_{B} / n_{C}\) of the refractive indices of liquids \(B\) and \(C .\)

Short Answer

Expert verified
The ratio \( n_B/n_C \) is approximately 1.227.

Step by step solution

01

Understanding the Problem

We need to find the ratio of refractive indices \(n_B/n_C\). We know the critical angle for total internal reflection for liquid B is \(36.5^\circ\) and for liquid C is \(47.0^\circ\).
02

Formula for Critical Angle

The formula for the critical angle \( \theta_c \) when a light ray passes from a denser medium (liquid A) to a less dense medium (liquid B or C) is \( \theta_c = \text{arcsin}( \frac{n_L}{n_H} ) \), where \(n_L\) is the refractive index of the less dense medium and \(n_H\) of the more dense one.
03

Apply the Critical Angle Formula for Liquid B

For liquid B, the critical angle \( \theta_c = 36.5^\circ \). Thus, \( \frac{n_A}{n_B} = \sin(36.5^\circ) \), where \(n_A\) is the refractive index of liquid A.
04

Apply the Critical Angle Formula for Liquid C

Similarly, for liquid C, the critical angle \( \theta_c = 47.0^\circ \). Thus, \( \frac{n_A}{n_C} = \sin(47.0^\circ) \).
05

Solving for Ratio \( n_B/n_C \)

Divide the two equations: \( \frac{n_A}{n_B} = \sin(36.5^\circ) \) and \( \frac{n_A}{n_C} = \sin(47.0^\circ) \). Thus, \( \frac{n_B}{n_C} = \frac{\sin(47.0^\circ)}{\sin(36.5^\circ)} \).
06

Calculating the Sine Values

Calculate \( \sin(36.5^\circ) \approx 0.596 \) and \( \sin(47.0^\circ) \approx 0.731 \).
07

Final Calculation

Substitute these values into the ratio: \( \frac{n_B}{n_C} = \frac{0.731}{0.596} \approx 1.227 \).

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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 (often symbolized as "n") is a measure that describes how light propagates through a medium, such as air, water, or any transparent substance. It is a dimensionless number that indicates how much the speed of light is reduced inside the medium compared to in a vacuum. If the refractive index of a medium is high, it means light travels slower through it.

  • Refractive index of vacuum is 1 (since light travels fastest in vacuum).
  • Substances like water and glass have refractive indices greater than 1.
  • The index of refraction determines the amount of bending or refraction of light entering the medium.
When light enters from one medium to another with a different refractive index, it changes speed and direction. This bending of light is why objects appear "bent" when partially submerged in water, like a straw in a glass.

Understanding refractive indices is essential for solving problems involving light behavior at interfaces, such as determining whether and how total internal reflection occurs.
Critical Angle
The critical angle is a key concept when discussing the refraction and reflection of light. It is the angle of incidence above which total internal reflection occurs when light travels from a medium with a higher refractive index to one with a lower refractive index.

  • It is the smallest angle of incidence for which the refracted ray is completely reflected back into the medium.
  • Beyond this angle, no refraction occurs; light is entirely reflected.
  • The critical angle depends on the refractive indices of the two media.
When light passes from a denser medium like water (where it travels more slowly) to a less dense medium like air (where it travels faster), there is a specific angle—the critical angle—at which light will no longer exit the water but instead reflect back inside.

In the solution to the exercise, we calculate the critical angle using the formula \[ \theta_c = \text{arcsin} \left( \frac{n_L}{n_H} \right) \]where \( n_H \) is the refractive index of the denser medium and \( n_L \) is that of the less dense medium.
Snell's Law
Snell’s Law, also known as the law of refraction, defines the relationship between angles of incidence and refraction when light enters a different medium. It is expressed mathematically as:\[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \]where:
  • \( n_1 \) and \( n_2 \) are the refractive indices of the two media.
  • \( \theta_1 \) is the angle of incidence.
  • \( \theta_2 \) is the angle of refraction.
Snell's Law is pivotal in calculating how much a light ray bends, depending on the media it transitions through.

When the light passes through the interface between two media with different refractive indices, it is refracted toward the normal if it enters a denser medium and away from the normal if it moves to a less dense medium.

Using Snell's Law, we can derive the formula for the critical angle, particularly when dealing with total internal reflection scenarios. It provides the foundation for understanding the behavior of light as it travels through different substances and aids in calculations similar to those in the exercise provided.

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Most popular questions from this chapter

A ray of light is traveling in glass and strikes a glass–liquid interface. The angle of incidence is \(58.0^{\circ}\) , and the index of refraction of glass is \(n=1.50 .\) (a) What must be the index of refraction of the liquid so that the direction of the light entering the liquid is not changed? (b) What is the largest index of refraction that the liquid can have, so that none of the light is transmitted into the liquid and all of it is reflected back into the glass?

A scuba diver, submerged under water, looks up and sees sunlight at an angle of \(28.0^{\circ}\) from the vertical. At what angle, measured from the vertical, does this sunlight strike the surface of the water?

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?

A converging lens \((f=25.0 \mathrm{cm})\) is used to project an image of an object onto a screen. The object and the screen are 125 \(\mathrm{cm}\) apart, and between them the lens can be placed at either of two locations. Find the two object distances.

(a) For a diverging lens is \((f=-20.0 \mathrm{cm})\) construct a ray diagram to scale and find the image distance for an object that is 20.0 cm from the lens. (b) Determine the magnification of the lens from the diagram.

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