The refractive index of the material of an equilateral prism is √2. The angle of minimum deviation of that prism is:
30°
Understanding the Equilateral Prism Problem
This question asks us to find the angle of minimum deviation for an equilateral prism given its refractive index. An equilateral prism has a specific shape where all three angles are equal to 60°. The refractive index tells us how much the material of the prism bends light.
What is an Equilateral Prism?
An equilateral prism is a prism where all three angles are equal. The angle of the prism, denoted by \(A\), is the angle between the two refracting faces. For an equilateral prism:
- Prism angle \(A = 60^\circ\)
Given Information
We are given the refractive index of the material of the equilateral prism:
- Refractive index, \( \mu = \sqrt{2} \)
Finding the Angle of Minimum Deviation
When light passes through a prism, it deviates from its original path. The deviation angle changes with the angle of incidence. The minimum deviation angle, \( \delta_m \), is the smallest deviation angle possible for light passing through the prism. This occurs when the angle of incidence equals the angle of emergence, and the ray inside the prism is parallel to the base.
The relationship between the refractive index (\( \mu \)), the prism angle (\( A \)), and the angle of minimum deviation (\( \delta_m \)) is given by the prism formula:
\( \mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \)
Step-by-Step Calculation
Let's plug in the given values into the prism formula:
We have \( A = 60^\circ \) and \( \mu = \sqrt{2} \).
\( \sqrt{2} = \frac{\sin\left(\frac{60^\circ + \delta_m}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} \)
First, let's calculate the value of the denominator:
\( \sin\left(\frac{60^\circ}{2}\right) = \sin(30^\circ) \)
We know that \( \sin(30^\circ) = \frac{1}{2} \).
Substitute this back into the formula:
\( \sqrt{2} = \frac{\sin\left(\frac{60^\circ + \delta_m}{2}\right)}{1/2} \)
Now, we need to solve for \( \sin\left(\frac{60^\circ + \delta_m}{2}\right) \). Multiply both sides by \( 1/2 \):
\( \sin\left(\frac{60^\circ + \delta_m}{2}\right) = \sqrt{2} \times \frac{1}{2} = \frac{\sqrt{2}}{2} \)
Now we need to find the angle whose sine is \( \frac{\sqrt{2}}{2} \). We know that \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \).
Therefore,
\( \frac{60^\circ + \delta_m}{2} = 45^\circ \)
Multiply both sides by 2:
\( 60^\circ + \delta_m = 45^\circ \times 2 \)
\( 60^\circ + \delta_m = 90^\circ \)
Finally, solve for \( \delta_m \):
\( \delta_m = 90^\circ - 60^\circ \)
\( \delta_m = 30^\circ \)
So, the angle of minimum deviation for the equilateral prism with a refractive index of \( \sqrt{2} \) is \( 30^\circ \).
| Parameter | Symbol | Value |
|---|---|---|
| Prism Angle (Equilateral) | \(A\) | \(60^\circ\) |
| Refractive Index | \( \mu \) | \( \sqrt{2} \) |
| Angle of Minimum Deviation | \( \delta_m \) | To be calculated |
| Step | Calculation | Result |
|---|---|---|
| 1 | Prism Formula | \( \mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \) |
| 2 | Substitute A and \( \mu \) | \( \sqrt{2} = \frac{\sin\left(\frac{60^\circ + \delta_m}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} \) |
| 3 | Calculate \( \sin(A/2) \) | \( \sin(30^\circ) = 1/2 \) |
| 4 | Substitute \( \sin(A/2) \) | \( \sqrt{2} = \frac{\sin\left(\frac{60^\circ + \delta_m}{2}\right)}{1/2} \) |
| 5 | Solve for \( \sin\left(\frac{60^\circ + \delta_m}{2}\right) \) | \( \sin\left(\frac{60^\circ + \delta_m}{2}\right) = \sqrt{2} \times \frac{1}{2} = \frac{\sqrt{2}}{2} \) |
| 6 | Find the angle | \( \frac{60^\circ + \delta_m}{2} = 45^\circ \) |
| 7 | Solve for \( \delta_m \) | \( 60^\circ + \delta_m = 90^\circ \implies \delta_m = 30^\circ \) |
Revision Table: Key Prism Concepts
| Concept | Description | Formula |
|---|---|---|
| Prism Angle | Angle between the two refracting faces. | \(A\) |
| Deviation Angle | Angle between the incident ray and the emergent ray. | \( \delta = i + e - A \) |
| Minimum Deviation Angle | Smallest deviation angle, occurs when \(i=e\). | \( \delta_m \) |
| Prism Formula (at \( \delta_m \)) | Relates \( \mu \), \(A\), and \( \delta_m \). | \( \mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \) |
Additional Information: Factors Affecting Deviation
The angle of deviation of light passing through a prism depends on several factors:
- Angle of Incidence (\(i\)): The deviation angle decreases as the angle of incidence increases, reaches a minimum value (\( \delta_m \)), and then increases again.
- Prism Angle (\(A\)): A larger prism angle generally results in a larger deviation angle.
- Refractive Index (\( \mu \)): A higher refractive index of the prism material causes greater bending of light, leading to a larger deviation angle. The refractive index depends on:
- Material of the prism
- Wavelength of light (dispersion)
- Temperature
The minimum deviation angle (\( \delta_m \)) is a unique value for a given prism (material and angle) and wavelength of light.
Important Questions from Moving Charge and Magnetism
A square loop with each side 1 cm, carrying a current of 10 A, is placed in a magnetic field of 0.2 T. The direction of magnetic field is parallel to the plane of the loop. The torque experienced by the loop is:
A long straight wire of circular cross-section with radius ' a ' carries a steady current ' I ' which is uniformly distributed across the cross-section. The magnetic field in the region r < a and r > a is represented by:
A 300-turn rectangular coil of length 20cm and breadth 12cm carries a current of 12A in a magnetic field of 6T. The plane of the coil makes an angle of 60∘ with the magnetic field. What is the torque acting on the coil?
An alpha-particle moves with a speed of 5×105m/s. It enters a region where there is a magnetic field of magnitude 4 T, directed at an angle of 45° to the X-axis and lying in the XY plane. The magnitude of the magnetic force on the alpha-particle is:
An electron moves around the nucleus in a hydrogen atom of radius 0.05nm with a velocity of 2×106m/s. The magnetic field produced at the center of the nucleus is:
