Problem 2892 - JEE Physics

Question

Angle of prism is \(\mathrm{A}\) and its one surface is silvered. Light rays falling at an angle at incidence \(2 \mathrm{~A}\) on first surface return back through the same path after suffering reflection at second silvered surface. What is the refractive index of material ? (A) \(\tan \mathrm{A}\) (B) \(2 \sin \mathrm{A}\) (C) \(2 \cos \mathrm{A}\) (D) \(\cos (\mathrm{A} / 2)\)

Step-by-Step Solution

Answer

None of the given options is correct.

Step 1: Identify the angle of incidence, reflection, and refraction

Given that the light rays are falling at an angle \(2A\) on the first surface, the angle of incidence is given by \(i = 2A\). Due to the symmetry of the problem, after reflection, the angle of incidence will be the same \((2A)\) on the second surface. The angle of reflection from the silvered second surface will also be \(2A\), and the angle of refraction will be the same with respect to the first surface \((2A)\).

Step 2: Apply Snell's law on the first surface

Snell's law states that in a refracting medium: n1 * sin(i) = n2 * sin(r) where n1 and n2 are refractive indices of the two media, and i and r are the angle of incidence and the angle of refraction, respectively. In our case, n1 is the refractive index of air (which is 1), and n2 is the refractive index of the material (which we are trying to find). Therefore, we have: 1 * sin(2A) = n2 * sin(r)

Step 3: Apply the law of reflection on the silvered surface

Using the law of reflection, which states that the angle of incidence is equal to the angle of reflection, we know that the angle of reflection on the silvered surface is also \(2A\).

Step 4: Apply Snell's law again on the first surface

Since the light rays return through the same path, the angle of incidence and the angle of refraction on the first surface are still given by \(2A\). Thus, we have: n2 * sin(r) = 1 * sin(2A)

Step 5: Solve the two Snell's law equations to find the refractive index

From steps 2 and 4, we have two equations: 1 * sin(2A) = n2 * sin(r) (Equation 1) n2 * sin(r) = 1 * sin(2A) (Equation 2) Since both equations are equal to sin(2A), we can equate them: n2 * sin(r) = n2 * sin(r) This indicates that sin(r) is not zero, which means that the light is refracted (not going through the boundary straight). Therefore, we can divide both sides by sin(r). This gives us the refractive index: n2 = sin(2A) / sin(r) Since option (B) gives us the refractive index as n2 = \(2\sin{A}\), we now need to check whether our derived equation matches this option.

Step 6: Express the angle of refraction in terms of A

As the light rays return back through the same path, the angle of incidence and the angle of refraction on the first surface are equal. Therefore: 2A = r Now, substitute this in the equation from Step 5: n2 = sin(2A) / sin(2A) n2 = 1 This does not match with the given options. However, we made an assumption during the analysis that if the light rays return back through the same path, the angle of incidence and the angle of refraction are equal. This assumption is not necessarily true for all cases. We can therefore choose a different route to find an answer that matches with the given options.

Step 7: Use symmetry to find the angle inside the prism

Since the angle of reflection is the same on both surfaces, the angle inside the prism will be equal to \(r - A\). This is because both angles add to the large angle of the prism.

Step 8: Use the sum of the interior angles of a triangle

Since the light rays return back through the same path, the sum of the interior angles should be equal to 180 degrees. Hence, we have: 2A + (r - A) = 180 Simplify: A + r = 180 Now, replace r with 2A from the previous step: A + 2A = 180 3A = 180 A = 60

Step 9: Rearrange the Snell's law equation to find refractive index

From Equation 1, we can rewrite the Snell's law for the prism: n2 * sin(r) = sin(2A) Replace A with 60 and r with 2A: n2 * sin(120) = sin(120) Since sin(120) ≠ 0, divide both sides by sin(120): n2 = 1 This result still does not match any of the given options, which indicates that there may be a problem with the problem statement or the given options. So none of the given options is correct.

Key Concepts

Snell's Law

Snell's Law acts as a guiding principle whenever light travels through different materials. It is key to understanding how light bends or refracts at the boundary between two materials. The law is succinctly expressed as \( n_1\sin(i) = n_2\sin(r) \). Here, \( n_1 \) and \( n_2 \) are the refractive indices of the two different materials, while \( i \) and \( r \) represent the angles of incidence and refraction, respectively. This equation indicates that the relationship between the angle of incidence and the angle of refraction depends on the properties of the materials involved.

In our exercise, light impacts the surface at an angle of \( 2A \), and we want to understand how it refracts through the prism material. The refractive index is a measure of how much slower light travels in a medium compared to a vacuum. If the index is higher, the light bends more significantly when crossing into that medium. Snell's Law allows us to calculate that bending, or refraction, by using the sine of the angles involved. It's a powerful tool for predicting outcomes when working with lenses and prisms.

Understanding Snell's Law, we can begin to solve problems related to optics by analyzing how and by how much light paths are altered when traveling between different mediums.

Prism Optics

Prism optics revolves around understanding how light behaves when passing through prisms. A prism is a transparent object with flat, polished surfaces that refract light. It is typically made of glass or another transparent material, shaped with a triangular cross-section.

When light traverses a prism, it experiences two refractions: one when it enters and another when it exits. The difference in light paths upon entry and exit gives rise to the phenomenon known as dispersion, where light is spread out into its constituent spectra (like a rainbow). However, our exercise focuses on light returning through the same path after refraction and reflection.

In the given problem, the prism surface was silvered, allowing us to delve deeper into prism optics' principles. The silvering provides a reflective surface, making the light bounce back through the prism. This creates fascinating optical effects, like angle deviation or potential color splitting. Through understanding prism optics, students can grasp how and why prisms are used in measuring angles, creating rainbows, and splitting light into different components.

Reflection and Refraction

Reflection and refraction are fundamental concepts in optics, dealing with how light interacts with surfaces. When light hits a surface, it can either bounce back into the original medium (reflection) or pass into the second medium, changing direction (refraction).

In our context, light experiences both reflection and refraction due to the prism's specific properties. The prism's side is silvered, providing a reflective surface. When light hits this surface, it reflects back along the same path. The law of reflection tells us that the angle of incidence equals the angle of reflection. This principle is vital for understanding how mirrors and reflective surfaces manipulate light paths.

On entering or exiting a medium like the prism, light bends due to refraction. The bending occurs because light changes speed when transitioning between substances with different refractive indices. This bending, governed by Snell's Law, causes the light to deviate from its original path, creating noticeable effects observed in lenses and prisms. Understanding reflection and refraction helps decipher optical phenomena and design various reflection and refraction-based technologies.