Question

There is a prism with refractive index equal to \(\sqrt{2}\) and the refracting angle equal to \(30^{\circ} .\) One of the refracting surfaces of the prism is polished. A beam of monochromatic light will retrace its path if its angle of incidence over the refracting surface of the prism is \(\ldots \ldots\) (A) \(45^{\circ}\) (B) \(0^{\circ}\) (C) \(60^{\circ}\) (D) \(30^{\circ}\)

Short answer

In order to find the angle of incidence for which a beam of monochromatic light will retrace its path after reflection on one of the refracting surfaces of a prism with refractive index equal to √2 and refracting angle equal to 30°, we need to apply Snell's Law and find the critical angle. By considering the condition for total internal reflection, we find the critical angle to be 45°. Therefore, the angle of incidence must be equal to the critical angle, which is also 45°. The correct choice is (A) \(45^{\circ}\).

Step-by-step solution

Understand the problem

We are given a prism with a refractive index equal to √2 and the refracting angle equal to 30°. One of the refracting surfaces is polished, and we are asked to find the angle of incidence for which a beam of monochromatic light will retrace its path after reflection.

Apply Snell's Law

When light enters the prism, it refracts and follows Snell's Law. According to Snell's law, when light passes from one medium to another, the product of the refractive index and the sine of the angle of incidence (in medium 1) is equal to the product of the refractive index and the sine of the angle of refraction (in medium 2). Mathematically, it can be expressed as: \(n_1 \sin{i_1} = n_2 \sin{i_2}\). Here, \(n_1\) and \(i_1\) are the refractive index and angle of incidence in medium 1, and \(n_2\) and \(i_2\) are the refractive index and angle of refraction in medium 2.

Find the critical angle

To find the critical angle, we need to consider the condition for total internal reflection. For this condition, the angle of refraction should be equal to 90°. Thus, applying Snell's law when the beam of light leaves the prism: \(n_{prism} \sin{i_c} = n_{air} \sin{90^{\circ}}\), where \(i_c\) is the critical angle and \(n_{prism}\) and \(n_{air}\) are the refractive indices of the prism and air, respectively. Since the refractive index of air is approximately 1, we can write: \(\sqrt{2} \sin{i_c} = 1\). Solving for \(i_c\), we get: \(\sin{i_c} = \frac{1}{\sqrt{2}}\). Therefore, \(i_c = 45^{\circ}\).

Determine the angle of incidence

In order for the light beam to retrace its path, it needs to internally reflect at the critical angle. Therefore, the angle of incidence must be equal to the critical angle. So, the angle of incidence is \(i = i_c = 45^{\circ}\). Thus, the correct choice is: (A) \(45^{\circ}\).

Key concepts

Refractive Index

The refractive index is a measure of how much the speed of light is reduced inside a particular medium compared to its speed in a vacuum. Light changes speed when it travels through different materials, and the refractive index quantifies this effect.

• A high refractive index indicates that light travels slower in the medium compared to in a vacuum.
• It is calculated as the ratio of the speed of light in a vacuum to the speed of light in the substance.
• For example, in the original exercise, the prism has a refractive index of \(\sqrt{2}\).

When light enters a medium with a different refractive index, it bends or refracts. Snell's Law helps predict the degree of this bending by relating the refractive indices of two media to the angles of incidence and refraction. If you imagine a ray of light entering the prism in the exercise, the refractive index plays a key role in determining how that light ray bends and the angles involved.

Critical Angle

The critical angle is an important concept when discussing refraction and reflection in optics. It is defined as the angle of incidence above which light cannot pass through the interface between two different media and instead reflects entirely back into the original medium.

• This phenomenon occurs when light moves from a medium with a higher refractive index to one with a lower refractive index, such as from glass to air.
• At the critical angle, the angle of refraction equals \(90^{\circ}\), meaning the refracted ray runs along the boundary.
• In the exercise, the critical angle is calculated using Snell's Law: \( \sqrt{2} \sin {i_c} = 1 \), where solving this gives \( i_c = 45^{\circ} \).

Understanding the critical angle is crucial for predicting behaviors such as the complete reflection of light within a medium, which has applications in fiber optics, binoculars, and more.

Total Internal Reflection

Total internal reflection is a fascinating optical phenomenon that occurs when the angle of incidence is greater than the critical angle. At this point, all the light is reflected back into the original medium instead of being refracted out of it.

• This phenomenon is used in devices like fiber optic cables, which rely on total internal reflection to transmit data over long distances without loss.
• In our prism exercise, total internal reflection is what allows the light beam, upon reaching the critical angle, to retrace its path.
• It is a reflection efficiency that is almost impossible to achieve with ordinary mirrors.

Total internal reflection ensures that light can be directed with precision and efficiency, highlighting how manipulating the angle of incidence via the critical angle results in remarkable optical applications.