Problem 18 - Physics

Question

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 \(\mathrm{A}\) with a \(30.0^{\circ}\) angle of incidence. At what angle of refraction does the ray leave the glass at point B?

Step-by-Step Solution

Answer

The light ray exits the glass with an angle of refraction of approximately \( 29.9^{\circ} \).

Step 1: Understanding the Problem

To find the angle at which the light ray exits the glass, we need to use Snell's Law at both surfaces of the glass. Snell's Law is given by: \[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \] where \( n_1 \) 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.

Step 2: Applying Snell's Law at the First Interface

For the light entering the glass, we have:- \( n_1 = 1.63 \) (carbon disulfide)- \( n_2 = 1.52 \) (glass)- \( \theta_1 = 30.0^{\circ} \)Applying Snell's Law:\[ 1.63 \sin(30.0^{\circ}) = 1.52 \sin(\theta_2) \]Solving for \( \theta_2 \):\[ \sin(\theta_2) = \frac{1.63 \times 0.5}{1.52} \approx 0.536 \]\[ \theta_2 \approx \sin^{-1}(0.536) \approx 32.3^{\circ} \]

Step 3: Finding the Exit Angle at the Second Interface

Now, applying Snell's Law again at the second glass-liquid interface:- \( n_1 = 1.52 \) (glass)- \( n_2 = 1.63 \) (carbon disulfide)- \( \theta_3 = 32.3^{\circ} \) (from the refraction inside the glass)Applying Snell's Law:\[ 1.52 \sin(32.3^{\circ}) = 1.63 \sin(\theta_4) \]Solving for \( \theta_4 \):\[ \sin(\theta_4) = \frac{1.52 \sin(32.3^{\circ})}{1.63} \]\[ \sin(\theta_4) \approx \frac{1.52 \times 0.536}{1.63} \approx 0.499 \]\[ \theta_4 \approx \sin^{-1}(0.499) \approx 29.9^{\circ} \]

Step 4: Concluding the Solution

The angle of refraction at which the ray exits the glass block into the liquid carbon disulfide is approximately \( 29.9^{\circ} \). Use a calculator for precise sine and inverse sine calculations to ensure accuracy.

Key Concepts

Refractive Index

The refractive index of a material is a measure of how much light bends as it enters the material from another medium. Think of it as the optical density of a material. The refractive index is denoted by the symbol \( n \). It is calculated as the ratio of the speed of light in vacuum to the speed of light in the medium.
It represents how quickly or slowly light can travel through a substance compared to empty space. A higher refractive index means that light travels slower in that medium.
In our example, the refractive index of glass is \( n = 1.52 \) while carbon disulfide has a higher refractive index of \( n = 1.63 \). This indicates that light travels slower in carbon disulfide than in glass.

Angle of Incidence

The angle of incidence is the angle that a ray of light makes with the normal (an imaginary line perpendicular) at the point where the light hits the surface of a material. This angle is crucial in determining how light will behave when it hits the interface between two different media.
In Snell's Law, the angle of incidence is denoted as \( \theta_1 \). It is used along with the refractive index to calculate the angle of refraction when light moves from one medium to another.

• In our problem, the light ray hits the glass at an angle of incidence of \( 30.0^{\circ} \).

Understanding this angle helps predict how and where light will travel next once it interacts with a new surface.

Angle of Refraction

The angle of refraction is the angle between the refracted ray and the normal at the point where the ray enters the new medium. Snell's Law allows us to calculate this angle, as it describes how light bends when it travels from one medium to another.
This is represented by \( \theta_2 \) or \( \theta_4 \) in Snell's equation, depending on which interface is being observed.

• In the exercise, the angle of refraction into the glass is found to be \( 32.3^{\circ} \).
• When the light exits the glass block, the angle of refraction changes again, calculated to be approximately \( 29.9^{\circ} \) into the carbon disulfide.

By understanding the angle of refraction, we can predict and explain the bending paths that light rays take, which is essential in many practical applications like lens design and fiber optics.