A beam of light in water \(\left(n_1=4 / 3\right)\) strikes an interface with a piece of glass \(\left(n_2=1.5\right)\). The critical angle at which total internal reflection takes place is (A) \(0^{\circ}\) (B) \(48.6^{\circ}\) (C) \(62.7^{\circ}\) (D) \(90^{\circ}\) (E) Total internal reflection cannot take place

Short Answer

Expert verified
Given that the refractive index of water is 4/3, and the refractive index of glass is 1.5. A) Light is completely reflected B) Light is completely refracted C) There is no critical angle D) Light both reflects and is refracted E) Total internal reflection cannot take place

Step by step solution

01

Write down Snell's Law for the given situation

We can define Snell's Law as \(\frac{\sin(\theta_1)}{\sin(\theta_2)} = \frac{n_2}{n_1}\). Given that \(n_1 = 4/3\), \(n_2 = 1.5\), and total internal reflection takes place when \(\theta_2 = 90^{\circ}\), we have to determine the critical angle \(\theta_1\).
02

Solve for the critical angle

We know that \(\theta_2 = 90^{\circ}\), so \(\sin(\theta_2) = \sin(90^{\circ}) = 1\). Our equation from Snell's Law now becomes: \(\frac{\sin(\theta_1)}{1} = \frac{1.5}{4/3}\) To solve for \(\theta_1\), we need to find the inverse sine of the right-hand side of the equation. First, simplify the right-hand side: \(\frac{1.5}{4/3} = \frac{1.5}{1} \times \frac{3}{4} = \frac{9}{8}\) Now, find the inverse sine: \(\theta_1 = \arcsin(\frac{9}{8})\) Since the sine function has a maximum value of 1, there is no value of \(\theta_1\) that satisfies this equation. This means that total internal reflection cannot take place.
03

Choose the correct answer

The correct answer is (E) Total internal reflection cannot take place.

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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 optics, vital for understanding how light bends, or refracts, as it passes from one medium into another. In simple terms, this law connects the ratio of the sines of the angles of incidence and refraction to the ratio of the indices of refraction of the two media. Mathematically, Snell's Law is expressed as:
\[\begin{equation}\frac{\sin(\theta_1)}{\sin(\theta_2)} = \frac{n_2}{n_1}\end{equation}\]
where \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction, and \(n_1\) and \(n_2\) are the refractive indices of the first and second medium, respectively.

Applying Snell's Law

To apply Snell's Law, one must identify these angles and indices. As light enters a new medium, it bends towards the normal if moving to a more optically dense medium (higher refractive index), or away from it if moving to a less optically dense medium (lower refractive index).

Limitations of Snell's Law

It's important to note that Snell's Law assumes that the light enters the second medium at all—this is not always the case, as with total internal reflection. When light moves from a medium with a higher refractive index to one with a lower index, there exists an angle of incidence beyond which refracted light is not possible; instead, the light is entirely reflected back into the medium, a phenomenon Snell's Law helps to predict.
Critical Angle
The critical angle is a key term when discussing total internal reflection. It refers to the specific angle of incidence at which light, traveling from a medium with a higher refractive index to one with a lower refractive index, refracts along the boundary, producing an angle of refraction of 90°.

Calculating the Critical Angle

One can calculate the critical angle using Snell's Law by setting \(\theta_2\) to 90°, indicating that the refracted ray is grazing along the interface. From Snell's Law, we get:\[\begin{equation}\sin(\theta_{critical}) = \frac{n_2}{n_1}\end{equation}\]For light to undergo total internal reflection, the angle of incidence \(\theta_1\) must be greater than the critical angle.
An interesting aspect of the critical angle is that it is apparent only when light is trying to move from a denser to a rarer medium (like from water to air). Moreover, when the critical angle is exceeded, all the light is reflected back into the original medium, and this is what is known as total internal reflection—a phenomenon widely used in fiber optics.
Optics
Optics is the branch of physics that involves the study of light and its interactions with matter. It encompasses a wide range of phenomena including reflection, refraction, dispersion, diffraction, and polarization.

Importance of Optics

Optics is crucial in a variety of technological applications, such as lenses, microscopes, telescopes, and fiber optic communications. For students learning about optics, it is fundamental to understand how light behaves in various situations, particularly in transitioning between different media.

Reflection and Refraction

In the context of our exercise, two primary concepts of optics are involved: reflection and refraction. Reflection happens when light bounces off the surface of a material, while refraction is the change in direction of light as it passes from one medium to another. Both of these phenomena are governed by principles that include Snell's Law and the concept of the critical angle, which help explain various optically-related occurrences, such as the shimmering effect of a pool or the functionality of optical fibers in communication technologies.

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Most popular questions from this chapter

A beam of light traveling through a piece of glass emerges into air. Which of the following statements is true as the light exits the glass? i. The speed of light stays the same; ii. The light speeds up; iii. The light slows down; iv. The frequency of the light stays the same; v. The frequency of the light increases; vi. The frequency of the light decreases; vii. The wavelength of the light stays the same; viii. The wavelength of the light increases; ix. The wavelength of the light decreases. (A) i, iv, vii (B) i, v and vii (C) ii, iv and viii (D) ii, v, ix (E) iii, iv and ix

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An object initially at infinity is brought closer to the focus of a concave mirror. As the object moves in, (A) the inverted image moves closer to the focus and becomes larger. (B) the inverted image moves farther out from the focus and becomes larger. (C) the upright image moves closer to the focus and becomes larger. (D) the upright image gets farther out from the focus and becomes larger. (E) the inverted image gets farther from the focus and becomes smaller.

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