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Welcome to Physicscalculations.com! I’m Abbas Nura, a passionate and dedicated physics educator, and I’m excited to share my love for physics with you through this platform. My goal is to make physics accessible and enjoyable for all students, helping them excel in their studies and develop a deep understanding of the subject.
With a wealth of experience in both education and health advocacy, I bring a unique perspective to the world of physics education. I started my journey in 2010, working with the Health Education Initiative for Women (HEIFOW) immediately after graduating from Kano University of Science and Technology (KUST) Wudil, where I earned my bachelor’s degree in physics. This experience opened my eyes to the power of education in empowering individuals, particularly children, teenagers, women and men, to make informed choices about physics.
In 2012, I ventured into teaching physics, and it has been a truly rewarding experience. I firmly believe that physics is not as challenging as it may seem, and my passion for the subject drives me to create engaging and effective learning experiences. As a result, an impressive 90% of my students consistently pass my lessons with top grades, often achieving A’s in their exams.
To enhance my teaching skills and knowledge, I pursued a Postgraduate Diploma from the National Teachers Institute of Nigeria (NTI). This program equipped me with advanced teaching techniques and pedagogical strategies, enabling me to tailor my lessons to meet the diverse needs of my students effectively.
Fueling my thirst for knowledge, I obtained a Master’s degree in Theoretical and Mathematical Physics from the University of Abuja. This rigorous program expanded my understanding of physics and honed my analytical and problem-solving skills. It allows me to delve into the intricate theories and concepts that underpin the laws of the universe.
As a registered professional with the Teachers Registration Council of Nigeria (TRCN), I am committed to maintaining the highest standards of professionalism and ethics in the teaching profession. I continuously strive to stay updated with the latest advancements in physics education, ensuring that my students receive the most relevant and comprehensive instruction.
Physicscalculations.com is your go-to resource for all things physics, especially the calculations part. Whether you are a student seeking additional support, a teacher looking for innovative teaching methods, or an enthusiast fascinated by the wonders of the universe, this platform is designed to cater to your needs. Here, you will find a rich collection of tutorials, problem-solving exercises, and engaging discussions to help you grasp complex concepts and strengthen your understanding of physics.
I invite you to join our vibrant community of learners, where we share a common passion for physics. Let’s embark on a journey of exploration, collaboration, and growth together. Through this website, I aim to foster a supportive and interactive environment that nurtures learning and encourages curiosity.
When it comes to studying physics, there are two important concepts that every student must know: Vectors and Scalars. Both of these terms are used to describe the physical quantities. And those quantities help us to measure various aspects of motion, energy, and force.
Vectors and Scalars
However, understanding the differences between them is essential for mastering the subject and excelling in exams. In this article, we’ll dive into the details of vectors and scalars, their differences, and real-world applications.
What are Vectors?
A vector is a quantity that has both magnitude and direction. In other words, it’s a physical quantity that requires a direction to be fully defined. For instance, velocity is a vector quantity. This is because it describes the rate at which an object moves in a specific direction. They are represented by an arrow with a specific length and a specified direction. Vectors can describe physical quantities such as displacement (e.g., 10 meters east), velocity (e.g., 20 meters per second north), and force (e.g., 50 Newtons to the right). When working with vectors, you need to consider both the magnitude and direction, making vector mathematics more complex than scalar mathematics.
Similarly, force is also a vector quantity because it has both magnitude and direction. Some other examples of vector quantities are displacement, acceleration, and momentum.
What are Scalars?
On the other hand, scalars are quantities that have only magnitude but no direction. In other words, they’re the physical quantities that can be completely defined by their magnitude alone. Examples of scalar quantities include temperature, mass, time, and energy.
They are described by a single numerical value and are accompanied by a unit of measurement. Scalars don’t have a specific direction associated with them. Examples of scalars include mass (e.g., 5 kg), time (e.g., 10 seconds), and temperature (e.g., 30 degrees Celsius). When you perform mathematical operations on scalars, like addition or multiplication, you treat them as ordinary numbers.
Key Differences between Vectors and Scalars
One of the main differences between vectors and scalars is that vectors require direction while scalars do not. Also, we represent vectors by arrows to show their direction. Whereas we identify scalars by a single numerical value.
For instance, if we say that a car is moving at 50 miles per hour. Then the direction of the movement of the car is a scalar quantity. This is because it only describes the magnitude of the car’s speed but not its direction.
However, if we say that a car is moving at 50 miles per hour towards the east. We can then describe it as a vector quantity. This is because it includes both the magnitude and direction of the car’s velocity.
Another difference between vectors and scalars is how we add or subtract them. When adding or subtracting vectors, it’s important to consider both their magnitudes and directions. However, when adding or subtracting scalars, we only need to add or subtract their magnitudes.
For example, if you’re driving a car at 40 miles per hour, and then you increase your speed to 60 miles per hour, the change in speed is 20 miles per hour (a scalar quantity).
However, if you’re driving at 40 miles per hour towards the east and then change your direction to the north while increasing your speed to 60 miles per hour, you’ve added a vector quantity to your original velocity.
Applications of Vectors and Scalars
The concepts of vectors and scalars are not just limited to physics textbooks but have numerous real-world applications. For instance, we apply vectors in navigation and mapping. This is because they help in calculating the distance and direction between two points.
We also apply these quantities in engineering and design to determine the force required to move an object in a specific direction. On the other hand, we apply scalars in various fields such as finance, economics, and medicine to calculate and analyze numerical data.
Summary
Summarily, vectors and scalars are two essential concepts in the study of physics, and understanding their differences is crucial for excelling in the subject. While vectors require direction, scalars do not, and we add or subtract the two types of quantities differently.
Both vectors and scalars have practical applications in various fields, and being able to distinguish between them is necessary for understanding their real-world implications
When we talk of impedance vs resistance we are simply referring to the differences and similarities that exist between impedance vs resistance. In this article, we will look at what resistance, impedance are.
impedance vs resistance
We will also explain their concept, the formula applicable to each. Mention the differences between impedance vs resistance, and also have a look at their similarities. We will also solve a few problems on impedance and resistance.
What is a Resistance?
Resistance is the opposition to the flow of current or charges (electrons). The s.i unit of resistance is ohm. We can write the symbol for ohm as Ω. Resistance takes place in a direct current (D.C) circuit.
The unit of resistance is named after a scientist George Simon Ohm. George Ohm investigates the relationship between the potential difference between the edges of a wire and the current flowing through the wire.
The above introduction will now lead us to the explanation of ohm’s law to better understand the concept of resistance.
What is Ohm’s Law?
Ohm’s law stated that the current flowing through a conductor, be it a wire, is directly proportional to the voltage (potential difference) across its ends, provided other physical properties of the conductor remain constant.
The mathematical formula for the above law is
I ∝ V
Which can be rewritten as
I = VR
Where R is the resistance and a proportionality constant.
Base on the formula I = VR, we can make R subject of the formula such that
R = V / I
From the above formula
R is the resistance and its unit is in ohm (Ω)
V is the voltage and its unit is in volts (v)
I is the current and its unit is in ampere (A)
We can also say that resistance is the ratio of potential difference to current flowing in the circuit. Ohm’s law is not obeyed by such conductors as radio valves (e.g diodes), transistors, rectifiers, and gases.
What is Impedance?
Impedance (Z) is the overall opposition of a mixed circuit containing a resistor, an inductor and/or a capacitor. We also measure impedance in ohms (Ω). Unlike resistance, Impedance takes place in an alternating current (a.c) circuit.
Like I mentioned earlier, the s.i unit of impedance is in ohms
Therefore, we can say that the potential difference is
V = IZ
Which shows that the impedance is
Z = V / I
We can equally say that impedance is a measure of how much a circuit resist the flow of alternating current. Impedance consist of resistance, inductance, and capacitance. Also, the voltage and current in an impedance represents sine wave.
Differences: Impedance vs Resistance
The major difference between impedance and resistance are:
Impedance
Resistance
– Is the measure of the total opposition of the current flow in a circuit that includes both resistance and reactance – The symbol of impedance is Z – Resist the flow of alternating current in a circuit
– Its a measure of how much a material or component opposes the flow of electric current through it – The symbol of resistance is R – Resist the flow direct current in a circuit
Differences: Impedance vs Resistance
Similarities Between Impedance and Resistance
Here are the similarities between impedance and resistance:
The s.i unit of impedance and resistance is in ohms (Ω)
Both impedance and resistance are due to the resistance of the flow of current.
They play an important role in the design of electronic circuits.
They are both the ratio of current to voltage
Solved Problem
An alternating current supply of 240 volts is connected across a capacitor of 5 micro faraday and resistance of 50 ohms in series at 50 Hz. Find
a. Capacitive reactance
b. Impedance
Solution
a. Capacitive reactance = Xc = 1 / 2πfC = 1 / (2 x 3.142 x 50 x 5 x 10-6)
From the diagram above, a weight W = 60N is suspended horizontally by a string attached to a wall. Find the tension T on the string.
Answer
The answer to the above question for the tension of the string is 40√3.
Data
Weight, W = 60N
Tension, T = ?
Angle, θ = 300
Formula
Since W = Tcosθ
We will apply the formula below to solve the problem
T = W / cosθ
Solution
We will insert our data into the formula
Tension, T = W / cosθ = 60 / cos300
Since cos300 = √3 / 2
Thus
T = 60 / ( √3 / 2 ) = 60 x (2 x √3) = 120√3
And 120√3 = 69.828 = 40√3
Therefore
T = 40√3
Thus, the tension of the string is 40√3.
Hence, the solution to the question “From the diagram above, a weight W = 60N is suspended horizontally by a string attached to a wall. Find the tension T on the string” is 40√3.
The symbol vo is correctly written as V0 or v0. We write the zero sign as a subscript to v. In any physics formula, v0 represents an object’s initial or original speed in a specific direction. When the speed of an object is in a specific direction, we call it velocity.
The velocity of an object is defined as the displacement over time. We can write the definition of displacement in a mathematical form as
v = s / t
Where v = velocity
s = displacement [Note: Displacement is the measure of separation between two points in a specified direction. It’s different from a distance]
t = time taken
Formulae Where v0 is Involved
a = (v – v0) / t [a = acceleration, t = time, v = final velocity, and v0 = initial velocity]
v = v0t + at
s = v0t + (1/2)at2
v2 = v02 + 2as
The zero sign is to help you differentiate between two similar velocities, where one is the velocity of the object when it starts moving (v0). While the other is the final velocity of the object when it stops moving (v).
You can change it to v1 (initial velocity for v0) and v2 (final velocity for v) as long as it will help you understand the problem you are trying to solve. In some books, you will see u instead of v0 as the initial velocity of an object.
Therefore, the correct answer to the question “what does vo mean in physics?” is an object’s initial or original speed (velocity) in a specific direction
The dual nature of light is the capacity of light to exhibit both wave-like and particle-like behaviour. This duality arises from compelling experiments that demonstrate light’s ability to exist as both a particle and a wave. Additionally, the dual nature of light explains the ability of light to behave as a wave or as a particle. This is due to valid experiments that show light can behave as a particle, and at the same time it can also behave as a wave. Experiments like electron diffraction indicate that light can exist as a wave. While the photoelectric effect and Compton effect experiments show that light has the nature of a particle.
What is Light in Physics?
Light is defined as electromagnetic radiation that is noticeable to the human eye. When you raise your head up, you will be able to see the ever-shining sky. We use our eyes to watch our favourite movies. With our eyes, we can be able to play chess game. The knowledge of the behaviour of light helps us to see and cherish the world around us. Therefore, we can simply explain the dual nature of light as the ability of light to behave like a particle or like a wave.
This understanding of optics not only enriches our everyday experiences but also plays a vital role in enhancing vision. Think about the various optical devices that help us to see, from sunglasses protecting us from the sun’s rays to prescription eyeglasses tailored to our unique needs. These inventions owe their existence to our comprehension of light.
We also have optical devices like microscopes to assist in viewing very tiny objects, and cameras to capture memorable moments in our lives. Light is very special because it helps us to see and identify objects.
A wave is a disturbance that travels through a medium and transfers energy from one point to another, without any permanent displacement of the medium itself.
A wave motion is a process of transferring a disturbance (in form of kinetic energy) from one point to another in a medium without any transfer of the particles of the medium.
What is a Particle in Physics?
In physics, we can say that particle is a localized object that shows certain physical properties such as mass, charge, density, spin, or volume. Particles are the foundation of matter.
We know that matter is anything that has mass and can occupy space. Matter consists of three states which are solids, liquids, and gas. These states of matter were made up of particles.
There are two categories of particles and they are:
Elementary particles: These particles are indivisible and cannot be broken down into smaller particles. For example, some of the elementary particles are quarks, lepton, and gauge bosons.
Composite particles: They are made up of two or more particles. Examples of composite particles include atoms (protons, neutrons, electrons), and mesons (quarks and antiquarks).
Fundamental forces such as gravity, electromagnetic force, weak force, and strong force help particles to interact with each other. The interactions of these particles are mediated by force carriers otherwise known as the exchange of particles.
Dual Nature of Light
What is Electron Diffraction?
Electron Diffraction is a result of the diffraction of a beam of light due to a crystal or other regular lattice structure.
What is the Photoelectric Effect?
The photoelectric effect is the process whereby electrons are emitted from the surface of a metal when light is shown on it. It’s the liberation of photons from the surface of the metal.
What is Compton’s Effect?
Compton’s scattering refers to the scattering of X-rays or gamma rays by matter, such as electrons in a material. It explains that when a single X-ray photon collides with a free electron, the electron recoils off as though it were a perfectly elastic sphere. The experiment was first carried out in 1923 by an American physicist known as Arthur Compton.
Explanation of the Dual Nature of Light
Here is a simple way to explain the dual nature of light. We all know that there are so many things we can do with a ball. We can either throw the ball in the air or roll the ball on the floor.
Now let us replace the ball with light and the two scenarios “of throwing and rolling” with waves and particles respectively.
When light behaves like a wave, it can travel through things like air, water, and even space. Additionally, light can also bear different colors like the rainbow.
Therefore, once light behaves like a particle, we refer to that light as photons. We cannot see photons with our eyes because of their size (very small).
Therefore, light has the nature of a wave according to electron diffraction. It can also behave as a particle according to the photoelectric effect.
The above statement explains the dual nature of light to exist as a wave or as a particle. Light can travel straight due to its particle nature and can also bend due to its wave nature.
Dual Nature of Matter and Radiation
The dual nature of matter and radiation is a fundamental concept in physics that suggests both particles and waves can characterize these entities. Matter, which constitutes everything with mass and occupies space, exhibits particle-like properties with distinct characteristics such as mass, charge, and volume. On the other hand, radiation, which includes electromagnetic waves like light, behaves like waves, displaying attributes such as frequency and wavelength.
This duality extends our understanding of the physical world, revealing that entities, from electrons to photons, can exhibit both wave-like and particle-like behaviour. The dual nature of matter and radiation plays a central role in modern physics, as it underpins phenomena like electron diffraction and the emission spectrum, providing profound insights into the behaviour of particles and waves at the quantum level. It serves as a cornerstone in comprehending the mysteries of the subatomic realm and shapes our ability to unravel the fundamental forces and behaviours governing the universe.
Photon
A photon is a tiny particle of light, the smallest unit of electromagnetic radiation. Think of it as a little packet of energy that light is made of. Photons have some unique qualities. First, they don’t have mass, which means they are incredibly light. Second, they move really fast, at the speed of light itself.
At the heart of this duality lies the photon, the fundamental particle associated with electromagnetic radiation. Photons are, in essence, the carriers of light, embodying its particle-like nature.
Photons are like messengers of light. When light shines on things, it’s actually a bunch of these photons coming together. They carry energy, and when they hit something, they can make it warm or even make it shine with its own light, like in a lightbulb.
Photons are also important in explaining the dual nature of light. In certain situations, light acts like a wave, and in others, it behaves like a stream of these tiny particles. The knowledge of photons helps us make sense of how light and other forms of electromagnetic radiation work, from radio waves to X-rays, and even the colours we see in a rainbow.
Emission Spectrum
An emission spectrum is a pattern of colourful lines of light produced by objects when they release energy. It helps identify what substances are made of by their unique colour code. The emission spectrum, which is a fascinating topic in quantum physics, further illuminates the dual nature of light. It encompasses the distinct wavelengths of light emitted by atoms and molecules when they transition from higher energy states to lower ones.
The emission spectrum offers a unique fingerprint for each element, unveiling the discreet and quantized nature of energy in the atomic world.
An emission spectrum is like a colorful fingerprint that tells us about the light given off by objects, like atoms or molecules, when they get excited. Imagine you have a special pair of glasses that can see different colors of light. When these objects heat up or get energy in some way, they start to glow with specific colors.
The emission spectrum is like a chart that shows exactly which colors they give off. It’s a bit like a secret code for scientists because each element or substance produces a unique pattern of colors. This helps us identify what things are made of, even if we can’t see them directly. Emission spectra are important in many areas, from understanding stars’ composition to analyzing materials in labs. So, they’re like a colorful language that helps us unlock the secrets of the universe.
If the refractive indices of glass and water are 1.5 and 1.3 respectively. What will be the critical angle when the angle of refraction in the water medium is 90 degrees?
Data
Refractive index of glass, ng = 1.5
The refractive index of water, nw = 1.3
Angle of refraction in the water medium, θw = 900
Unknown
Critical angle, C = ?
Formula
We will apply the formula
C = sin-1 (nw sinθw / ng)
Solution
We can now apply our data into the formula to obtain
C = sin-1 (nw sinθw / ng) = sin-1 (1.3 x sin900 / 1.5)
The above expression will become
C = sin-1 (1.3 x 1 / 1.5)
We will now have
C = = sin-1 0.9 = 640
Therefore, the critical angle is 64 degrees.
Problem 2
What is the critical angle for a light ray traveling in water with a refractive index of 1.33 that is incident on the surface of the water above which air with a refractive index of 1.00? Answer to the nearest degree.
Data
Refractive index of water, nw = 1.33
The refractive index of air, na = 1.00
Unknown
Critical angle, C = ?
Formula
We will apply the formula below to solve the problem
C = sin-1 (na / nw)
Solution
We will insert our data into the formula to get
C = sin-1 (na / nw) = sin-1 (1 / 1.33)
After dividing 1 by 1.33 we will obtain
C = sin-1 0.75 = 48.60 = 490
Therefore, the critical angle for the light ray is 49 degrees.
Problem 3
Light rays travel through a layer of kerosene floating on the surface of water that has a refractive index of 1.33. Light rays that are incident on the interface of kerosene and water at angles of 16.9° from the surface or less are totally internally reflected. What is the refractive index of the kerosene? Give your answer to two decimal places.
Data
The refractive index of water, nw = 1.33
Critical angle, C = 900 – 16.90 = 73.10
Unknown
The refractive index of the kerosene, nk = ?
Formula
We will use the formula
nk = nw / sinC
Solution
Insert your data into the above formula to obtain
nk = nw / sinC = 1.33 / sin73.10
We will now have
nk = 1.33 / 0.96 = 1.385 = 1.39
Therefore, the refractive index of the kerosene, nk is 1.39 in two decimal places.
Definition: The critical angle is the angle of incidence in the denser medium when the angle of refraction in the less dense medium is ninety degrees (900). The critical angle is a concept in optics that describes the angle of incidence at which light transitions from one medium to another with a higher refractive index, typically from a denser medium (e.g., glass or water) to a less dense one (e.g., air). When the angle of incidence surpasses the critical angle, light no longer refracts into the second medium but instead undergoes total internal reflection, staying within the denser medium.
The formula to calculate the critical angle (θc) is given by:
θc = sin-1(n2 / n1)
Where:
θc is the critical angle.
n2 is the refractive index of the denser medium.
n1 is the refractive index of the less dense medium.
This phenomenon has numerous practical applications, such as in fibre optics, where it allows for the transmission of light signals over long distances within the core of optical fibres, preventing signal loss.
At values above the critical angle, the total internal reflection occurs, that is, the refracted ray disappears and a strong reflected ray appears.
What is Critical Angle and its Formula?
A critical angle is the angle of incidence for which the refracted ray emerges tangent to the surface of the angle.
Therefore, the critical angle formula that can help us to find critical angle based on the question available to us are: θc = sin-1 (1 / n) or θc = sin-1 (na / ng). Where: θc = critical angle, n = refractive index.
The critical angle is a term in physics and optics to describe the angle of incidence of a ray of light at which it refracts at an angle of 90 degrees to the normal (the line perpendicular to the surface of the medium it is entering).
Let’s assume we allow a ray of light to move from one medium to another. For example, say air to water or from water to glass. This ray of light will change direction because of the change in the speed of light in any of the mediums.
Therefore, we refer to the change in the direction of the ray of light as refraction. The amount of refraction depends on the angle of incidence of the light ray and the refractive index of the two media.
Therefore, to explain the critical angle in the simplest term. We can say that it’s the angle of incidence at which the refracted ray of light is parallel to the boundary between the two mediaat 90 degrees. In other words, the angle of incidence results in the refracted ray being bent at a 90-degree angle from the normal.
We can calculate critical angle by applying Snell’s law. The law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the refractive indices of the two media.
sin i / sin r = sin θc / sin 900 = gna = 1 / ang = 1 / n =na / ng
i = Angle of incidence
r = Angle of refraction = 900
θc = Critical Angle
gna = Refractive index of the ray of light from glass to air
ang = Refractive index of the ray of light from air to glass
n = Refractive index of glass by definition
How to Derive Critical Angle Formula
When the angle of refraction is 90 degrees, the sine of the angle of refraction is equal to one (1). Therefore, the sine of the critical angle is similar to the reciprocal of the refractive index of the two media.
sin r = sin 900 = 1
and sin i = sin θc
Thus, sin i / sin r = sin θc / sin 900 = sin θc / 1 = sin θc = gna = 1 / ang = 1 / n = na / ng
As stated earlier, the critical angle formula can be written as: θc = sin-1 (1 / n) or θc = sin-1 (na / ng). Where: θc = critical angle, n = refractive index.
The critical angle is an important concept in optics. This is because it determines whether a light ray will be refracted or reflected when it encounters a boundary between two media.
If the angle of incidence is greater than the critical angle. The ray of light will be reflected back into the medium from which it came. This phenomenon is called total internal reflection. We use it in various optical devices, such as fiber optic cables, and prism-based cameras.
Relationship Between Refractive Index and Critical Angle
If the refractive angle is 900
By applying Snell’s law, the refractive index for glass – air boundary is: gna = (sin θc) / (sin 900)
Now, the refractive index from air to glass is: ang = (sin 900) / (sin θc)
But sin 900 = 1
Therefore, the refractive index from air to glass is: ang = 1 / (sin θc)
Total Internal Reflection
When a ray of light passes from an optically denser medium to an optically less dense medium, there exists a weak internal reflection and a strong refraction. But as the angle of incidence increases, the angle of refraction also increases at the same time.
The intensity of the reflected ray gets stronger and that of the refractive ray becomes weaker. When the angle of incidence exceeds the critical angle, there exists no refraction and a total reflection occurs. We refer to this phenomenon in physics as total internal reflection.
The conditions under which total internal reflection occurs are:
Light rays must travel through a denser medium to a less dense medium
The angle of incidence must exceed the critical angle.
The refractive indices of glass and water are 1.5 and 1.3 respectively. What will be the critical angle when the angle of refraction in the water medium is 900?
Data:
The refractive index of water, nw = 1.3
Refractive index of glass, ng = 1.5
Refractive angle, θ = 900
Unknown:
Critical angle, θc = ?
Formula:
θc = sin-1 (nw sin900)/ ng
Solution
sin θc = (nw sin900)/ ng = (1.3 x 1) / 1.5 = 0.866
θc = sin-1 (0.866) = 600
Table for Substances Their Critical Angles and Refractive Indexes
substance
Refractive index (n)
Calculation θc = sin-1 (1 / n)
Critical angle (θc)
Vacuum
1.00
θc = sin-1 (1 / 1)
900
Pure Water
1.33
θc = sin-1 (1 / 1.33)
48.750
Carbondioxide
1.0005
θc = sin-1 (1 / 1.0005)
88.190
Ice
1.31
θc = sin-1 (1 / 1.31)
49.760
Ethyl Alcohol
1.36
θc = sin-1 (1 / 1.36)
47.330
Quartz
1.46
θc = sin-1 (1 / 1.46)
43.230
Vegetable oil
1.47
θc = sin-1 (1 / 1.47)
42.860
Olive oil
1.48
θc = sin-1 (1 / 1.48)
42.510
Acrylic
1.49
θc = sin-1 (1 / 1.49)
42.160
Table Salt
1.51
θc = sin-1 (1 / 1.51)
41.470
Glass
1.52
θc = sin-1 (1 / 1.52)
41.140
Sapphire
1.77
θc = sin-1 (1 / 1.77)
34.400
Zircon
1.92
θc = sin-1 (1 / 1.92)
31.390
Cubic zirconia
2.16
θc = sin-1 (1 / 2.16)
27.580
Diamond
2.42
θc = sin-1 (1 / 2.42)
24.410
Gallium phosphide
3.50
θc = sin-1 (1 / 3.50)
16.600
Table For Substances their refractive indexes and critical angles
What is the critical angle for light traveling from crown glass (n = 1.52) into water (n = 1.33)?
Answer
The final answer to the above question is that the critical angle is 61 degrees (610)
Proof
Here is the method I applied to solve the problem:
Data
We will identify our data from the above question
The light travels from crown glass = Incident Light = ni = 1.52
It then passes through water = Refracted ray of light = nr = 1.33
According to the definition of the critical angle. When light passes through a denser to a less dense medium, it forms an angle of 900. Therefore, the angle of refraction, θr = 900
Unknown
The critical angle, θi = ?
Formula
We will apply the formula
nisinθi = nrsinθr
Solution
what is the critical angle for light traveling from crown glass (n = 1.52) into water (n = 1.33)?
Hence, insert the data into the formula
1.52 x sinθi = 1.33 sin900
Divide both sides by 1.52 to obtain
sinθi = (1.33 x sin900) / 1.52 = (1.33 x 1) / 1.52 = 1.33 / 1.52 = 0.875
The magnification of the ocular lens, also known as the eyepiece lens. It is a measure of how much an optical instrument, like a microscope or a telescope, enlarges the image of an object that you observe. It is a very important component of the total magnification of the instrument. In this article, we will concentrate on what is the magnification of the ocular lens and how to find it.
The eyepiece lens is the lens through which you directly view the magnified image. It is typically located at the top of the instrument, closest to your eye. The design of Ocular lenses is to further magnify the image produced by the objective lens or primary lens of the optical instrument.
The magnification of the eyepiece lens is usually marked on the lens itself, often as a single number, such as 10x. This number indicates the factor by which the eyepiece lens magnifies the image. In the case of a 10x eyepiece, it means the image appears ten times larger when viewed through the eyepiece compared to what your naked eye would see without any optical aid.
When combined with the magnification of the objective lens, the eyepiece lens’s magnification determines the total magnification of the optical instrument. Total magnification is calculated by multiplying the magnification of the objective lens and the eyepiece lens.
Understanding the magnification of the eyepiece lens is important because it directly affects the overall magnification of the instrument, allowing you to see fine details and structures in objects that would be otherwise difficult or impossible to observe with the naked eye. Scientists, researchers, and students often rely on the eyepiece’s magnification to study a wide range of subjects in fields such as biology, astronomy, and materials science.
Detail Explanation
The ocular lens is a component of a microscope or a telescope that magnifies the image produced by the objective lens.
Microscope picture
The focal length of the lens determines the magnification of an eyepiece. It is the distance between the lens and the point where the light coming from the objective lens converges into a focused image.
You can calculate its magnification by dividing the focal length of the eyepiece by the focal length of the objective lens.
The magnification of the ocular lens is the ratio of the total magnification of a microscope or a telescope to the objective lens.
In a compound microscope, the ocular lens or otherwise known as the eyepiece is 10x. To obtain the total magnification, you will need to multiply the ocular lens by the objective lens.
The typical magnification of an ocular lens ranges from 10x to 25x. But higher magnifications of up to 40x or 50x are also available.
A higher magnification ocular lens can provide greater detail and allow the observer to see smaller structures or objects. However, it reduces the field of view and can make it more difficult to see the entire specimen.
Therefore, you need to know that the total magnification of a microscope or a telescope is the product of the magnifications of the objective lens and the ocular lens.
The magnification of the objective lens ranges from 4x to 100x or more.It is also selected based on the size of the specimen and the level of detail required.
The objective lens provides the initial magnification of the image. While the eyepiece magnifies the image further to make it easier to observe.
Total magnification refers to the level of enlargement or amplification achieved when viewing an object through an optical instrument, such as a microscope or a telescope. It is a very important parameter that quantifies how much larger an object appears when observed through the optical system compared to when viewed with the naked eye. Understanding how to calculate total magnification will increase your knowledge of optics and its applications in laboratories. Therefore, we need to understand what a microscope is. A microscope, otherwise known as a compound microscope is a tool that helps us to see microscopic objects (very tiny objects) that we cannot see with the naked eye.
how to calculate total magnification
There are different designs of microscopes today. However, you need to know that a microscope consists of a rack, eyepiece (ocular), stage, eyepiece tube (eyepiece holder), condenser, aperture, objective lenses, knobs, nose piece, and light source.
how to calculate total magnification
How to Calculate Total Magnification
To calculate the total magnification of an optical system, you will need to multiply the magnification of each individual component in the system.
A sample video of microscope
In other words, total magnification is the product of all magnifications along the light path from the object to the observer’s eye. We can also refer to the observer’s eye as a recording medium.
For example, if you have a microscope with an eyepiece that magnifies 10x, and an objective lens that magnifies 40x. The total magnification would be 10x multiplied by 40x to obtain 400x.
Therefore,
Total magnification = eyepiece x objective lens
Understanding How to Calculate Total Magnification
Total magnification is typically expressed as a ratio or a factor, indicating how many times the image has been magnified in comparison to the actual size of the object. To calculate total magnification, you need to consider two factors:
Objective Lens Magnification: The objective lens is the lens closest to the object you are observing in the optical instrument. It provides the primary magnification. The magnification of the objective lens is usually marked on the lens itself, such as 4x, 10x, 40x, or 100x. This number represents the factor by which the objective lens enlarges the object.
Eyepiece (Ocular) Lens Magnification: The eyepiece lens, or ocular lens, is the lens through which you directly observe the magnified image. It contributes additional magnification. The eyepiece lens magnification is also marked, often as 10x.
To calculate the total magnification, you simply multiply the magnification of the objective lens by the magnification of the eyepiece lens:
Total Magnification = Objective Lens Magnification × Eyepiece Lens Magnification
For example, if you are using a microscope with a 40x objective lens and a 10x eyepiece lens, the total magnification would be:
Total Magnification = 40x (Objective) × 10x (Eyepiece) = 400x
So, in this case, the image of the object you are viewing through the microscope would appear 400 times larger than its actual size.
Example 1
Thus, assuming
objective lens = 10x
Then
Total magnification = 10x * 10x = 20x
Example 2
With the object’s lens as 4x
Total magnification = eyepiece (ocular) x object’s lens
Which implies that the total magnification = 10x * 4x = 40x
Note: It is important to know that the power of the eyepiece is usually 10x. Additionally, the total magnification can vary depending on the combination of lenses and other optical components used in the optical system. It’s also dependent on the distance between the components and the object under observation.
The magnification of objective lenses are 4x, 10x, and 40x.
Note
Total magnification is a very important parameter when using optical instruments. This is because it determines the level of detail you can observe and the size of objects you can effectively study. Additionally, It allows scientists, researchers, and students to understand the micro and macro worlds with precision and clarity, making it an essential concept in fields such as biology, materials science, and astronomy.
