two lasers are shining on a double slit, with slit separation d . laser 1 has a wavelength of d/20 , whereas laser 2 has a wavelength of d/15 . the lasers produce separate interference patterns on a screen a distance 4.20 m away from the slits.

Answer:

Kinetic energy is 1187.5J

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 3000 × 2

We have the final answer as

Hope this helps you

The force field F(p) = -(k/p^2)e represents an inverse square force law, where k is a constant and p is the radial distance from the origin. To find the angular frequency of polar vibrations around stable circular motion, we can consider the equation of motion for a particle in a circular orbit.

For circular motion, the centripetal force is given by Fc = mω^2p, where m is the mass of the particle, ω is the angular frequency, and p is the radial distance.Equating the centripetal force to the force field, we have:-
(k/p^2)e = mω^2p
Simplifying, we get:
k/p^3 = mω^2
Taking the square root and rearranging, we find:
ω = √(k/(mp^3))
This is the angular frequency of polar vibrations around stable circular motion.To show that this frequency is equal to the rotational angular frequency of circular motion, we can recall that in circular motion, the rotational angular frequency ω_rot = v/p, where v is the tangential velocity.Since the tangential velocity v = ω_rot * p, we can substitute this into the equation for ω:
ω = √(k/(m * p^3)) = √(k/(m * p^2)) * (1/p)
Using the relation v = ω_rot * p, we can rewrite the expression as:
ω = √(k/(m * p^2)) * (1/p) = √(k/(m * p^2)) * (1/(v/p))
Simplifying, we get:
ω = √(k/(m * p^2)) * (p/v) = v/√(m * k)
This shows that the angular frequency ω is equal to the rotational angular frequency ω_rot. Therefore, the angular frequency of polar vibrations around stable circular motion is equal to the rotational angular frequency of circular motion.

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Answer:

Potential energy is stored energy and the energy of position––gravitational energy. There are several forms of potential energy. Electrical Energy is the movement of electrical charges. Everything is made of tiny particles called atoms.

Answer:

the filling stops when the pressure of the pump equals the pressure of the interior air plus the pressure of the walls.

Explanation:

This exercise asks to describe the inflation situation of a spherical fultball.

Initially the balloon is deflated, therefore the internal pressure is equal to the pressure of the air outside, atmospheric pressure, when it begins to inflate the balloon with a pump this creates a pressure in the inlet valve and as it is greater than the pressure inside, the air enters it, this is repeated in each filling cycle, manual pump.

When the ball is full we have two forces, the one created by the external walls and the one aired by the pressure of the pump, these forces are directed towards the inside, but the air molecules exert a pressure towards the outside, which translates into a force. When these two forces are equal, the pump is no longer able to continue introducing air into the balloon.

Consequently the filling stops when the pressure of the pump equals the pressure of the interior air plus the pressure of the walls.

The size of the difference threshold being greater for heavier objects than for lighter ones illustrates Weber’s Law.

Weber's law is a principle in psychophysics that describes the relationship between the actual change in stimulus and the change in a subjective sensation of that stimulus. Weber's law states that the ratio of the difference threshold to the initial stimulus magnitude is constant across all levels of the initial stimulus magnitude, whether the stimulus is light, sound, or some other sensory input.

For instance, assume that you're sitting in a room with the lights turned off. Your companion turns on a flashlight, and you're suddenly able to see the objects in the room. If the flashlight's brightness is increased by 10%, you can detect the difference. But if the flashlight's brightness is increased by 10% from a brighter initial level, you might not notice the difference.

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Given:

The charge is q1 = 9.4 nC

The charge q2 = -2.31 nC

The distance between them is r = 4.94 cm

To find the electric field strength at the midpoint between two charges.

Explanation:

The electric field strength at the midpoint will be the sum of electric field strength due to q1 and q2.

The electric field strength can be calculated by the formula

Here, k is the electrostatic constant whose value is

The electric field strength due to the charge q1 is

The electric field strength due to the charge q2 is

The electric field strength at the midpoint will be

Thus, the electric field strength at the midpoint between the two charges is 173000 V/m

Answer: 18.81m/s^2

Explanation:

Given the following :

Height of building = 0. 1 km = 100m

Horizontal distance = 85m

Using the equation :

S = 1/2gt^2

And S = 100, g = 9.8m/s^2

100 = 0.5(9.8)(t^2)

100 = 4.9(t^2)

t^2 = (100 / 4.9)

t^2 = 20.408

t = 4.5175214

t = 4.52s

Therefore, initial velocity of ball in horizontal direction;

Using the equation:

S = ut + 0.5at^2

a in horizontal direction = 0

Therefore,

S = ut

85 = u × 4.52

u = (85 / 4.52)

u = 18.805

u = 18.81m/s

Explanation:

thermal energy refers to the energy contained within a system that is responsible for its temperature. Heat the flow of thermal energy

Answer:

33.33 m/sec

Explanation:

A baseball travels 200 metes in 6 seconds,

what is the baseball’s velocity?

use the formula: velocity = distance over time

where (d) distance = 200 m

and (t) time = 6 sec.

plugin values into the formula:

v = d / t

  = 200 m / 6 sec

  = 33.33 m/sec.

therefore, the baseball's velocity is 33.33 m/sec

With a total mass of 200 kg and a velocity of zero, an astronaut is floating in space while tossing a 5 kg ball at a velocity of 20 m/s. After hurling the ball, the astronaut's velocity is -0.04 m/s.

Momentum is referred to as an object's rate of motion. It is the result of the mass and the velocity of the object. Linear momentum is another name for this concept. The kilogram-meter-per-second (abbreviated as kgm/s) is the SI unit for momentum.

The astronaut and the ball together weigh 205 kg (200 + 5 kilogramme).

The astronaut's initial velocity, v1, is zero because they are at rest.

200 kilogrammes is the astronaut's mass (m1).

The ball weighs 5 kg/m2.

The ball's velocity, v2, is 20 m/s.

The astronaut-ball system experiences no outside forces, hence its momentum is conserved. Before tossing the ball, the system has no net momentum. As a result, after the ball is thrown, the system's overall momentum is also zero.

Let v represent the astronaut's speed after throwing the ball.

The momentum of the astronaut and ball system after the ball is thrown is given by

(200 kg) v + (5 kg) v (-20 m/s) = 0

200 v = 5 * 20

v = 5*20/200 v = 0.5 m/s.

Therefore, the astronaut's velocity after throwing the ball, in m/s is -0.04 m/s (-0.5 m/s in the backward direction).

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answers

81.6 pounds rounded or just 81.571 if it needs to be exact

a) Using Kepler's third law, the orbital period of planet B located at 1.0 AU from the sun can be calculated. b) Given an orbital period of 4 years for planet C, we can determine its average distance from the sun.

Kepler's third law states that the square of the orbital period (P) of a planet is proportional to the cube of its average distance (a) from the sun. Mathematically, it can be expressed as:

\(\[P^2 = a^3\]\)

Given that planet B is located at an average distance of 1.0 AU from the sun, we can substitute this value into the equation to solve for P:

\(\[P^2 = (1.0 \, \text{AU})^3\]\)

Taking the square root of both sides, we find:

\(\[P = \sqrt{(1.0 \, \text{AU})^3}\]\)

Evaluating the expression, we get:

\(\[P \approx 1.0 \, \text{year}\]\)

Therefore, the orbital period of planet B is approximately 1.0 year.

Similarly, using Kepler's third law, we can solve for the average distance (a) of planet C from the sun. We have the equation:

\(\[P^2 = a^3\]\)

Given an orbital period (P) of 4 years, we can substitute this value into the equation to solve for a:

\(\[(4 \, \text{years})^2 = a^3\]\)

Simplifying, we get:

\(\[16 \, \text{years}^2 = a^3\]\)

Taking the cube root of both sides, we find:

\(\[a = \sqrt[3]{16 \, \text{years}^2}\]\)

Evaluating the expression, we get:

\(\[a \approx 2.52 \, \text{AU}\]\)

Therefore, if planet C has an orbital period of 4 years, its average distance from the sun is approximately 2.52 AU.

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Answer:8.1g

Explanation:

Explanation:

you can find by using. volume = mass/ density of aluminium.

The correct options for the factors that contribute to the Earth's magnetic field, which protects us from solar wind, are option b. the liquid outer core and option f. the Earth's rotation.

The Earth's magnetic field is generated by the motion of molten iron and nickel in the liquid outer core of the Earth (Option b). This motion, known as convection, generates electric currents that produce the magnetic field.

The Earth's rotation (Option f) also plays a significant role. The rotation of the Earth causes the convection currents in the liquid outer core to circulate and amplify the magnetic field, creating a dynamo effect.

The Earth's revolution (Option a) around the Sun and its tilt on its axis (Option c) are not directly responsible for the generation of the Earth's magnetic field. They are factors that influence climate and seasons on Earth but do not contribute to the magnetic field.

The seasons (Option d) are related to the Earth's axial tilt and its orbit around the Sun but are not directly connected to the generation of the magnetic field.

The solid inner core (Option e) is not involved in generating the Earth's magnetic field. It is mainly composed of solid iron and nickel and does not have the fluid motion required for the generation of a magnetic field.

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Answer:

Explanation:

The mass of an object given it's force and acceleration can be found by using the formula

\(m = \frac{f}{a} \\ \)

f is the force

a is the acceleration

From the question we have

\(m = \frac{15}{5} = 3 \\ \)

We have the final answer as

Hope this helps you

Student loans are generally considered to be an "investment in your human capital." This is because they are intended to help students finance their education and obtain the skills and knowledge needed to increase their future earning potential and career opportunities.

Unlike other types of loans, such as personal loans or credit card debt, student loans are typically low-interest and may offer flexible repayment options based on income. By investing in their education, students can improve their long-term financial prospects and increase their ability to contribute to society.

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Answer:

15.7 ; 78.5

Explanation:

Given that

The mechanic use a screw driver to install 1⁄4-20 UNC bolt

And, the effort force is 5lb

We need to find out the mechanical advantage

And, the resistance force

As we know that

The Mechanical advantage of a screw = Circumference ÷ pitch

where,  

Circumference = pi × d

 pi = 3.142, D = diameter

So,  

Circumference = 3.142 × (1 ÷ 4)

= 0.785 in

Now

Pitch = 1 ÷ TPI

Here TPI (thread per inch) = 20

Pitch = 1 ÷  20 = 0.05

So,

Mechanical advantage = 0.785 ÷ 0.05

= 15.7

Now Resistance force if effort force is 5lb

We know that

Mechanical advantage = Fr ÷ Fe

Here

Fe = effort force, Fr = resistance force

15.7 = Fr ÷ 5

Fr = 15.7 × 5

= 78.5 lbs

Light bends away from the normal, because it's moving from higher to lower refractive index.

Same bend-direction as when it goes from water into air.

A. spreadsheet or computer script, we can calculate the value, which is 9.80651834 m/s^2. B. the value gives g(90°, 10,000) ≈ 9.81937375 m/s^2. C. the value gives g(45°, 0) ≈ 9.80666553 m/s^2. D. the value gives: g(-66.5°, 3,048) ≈ 9.83065116 m/s^2. E. the cosine function is an even function (cos(-θ) = cos(θ)), the value of g will be different for φ and -φ, except when φ = 0 (equator) where cos(0) = cos(0) = 1.

To compute the acceleration of gravity at the given locations, we'll substitute the values of latitude (φ) and height above mean sea level (Z) into the formula for g(φ, Z).

(a) Baseline Road (latitude 40° North), at an elevation of one mile above mean sea level:

Latitude (φ) = 40°

Height (Z) = 1 mile = 1,609.34 meters

Substituting these values into the formula, we have:

g(40°, 1609.34) = 980.6160(1 - 0.0026373cos^2(40°) + 5.9×10^(-6)cos^2(2 * 40°)) - (3.085462×10^(-4) + 2.27×10^(-7)cos^2(40°)) * 1609.34 + (7.254×10^(-11) + 1.0×10^(-18)cos^2(40°)) * (1609.34)^2 - (1.517×10^(-17) + 6×10^(-20)cos^2(40°)) * (1609.34)^3

Using a spreadsheet or computer script, we can calculate the value, which is approximately: g(40°, 1609.34) ≈ 9.80651834 m/s^2.

(b) 10 km above the North Pole:

Latitude (φ) = 90° (North Pole)

Height (Z) = 10,000 meters

Substituting these values into the formula, we have:

g(90°, 10,000) = 980.6160(1 - 0.0026373cos^2(90°) + 5.9×10^(-6)cos^2(2 * 90°)) - (3.085462×10^(-4) + 2.27×10^(-7)cos^2(90°)) * 10,000 + (7.254×10^(-11) + 1.0×10^(-18)cos^2(90°)) * (10,000)^2 - (1.517×10^(-17) + 6×10^(-20)cos^2(90°)) * (10,000)^3

Calculating the value gives: g(90°, 10,000) ≈ 9.81937375 m/s^2.

(c) Halfway between the equator and the North Pole at mean sea level:

Latitude (φ) = 45°

Height (Z) = 0 meters

Substituting these values into the formula, we have:

g(45°, 0) = 980.6160(1 - 0.0026373cos^2(45°) + 5.9×10^(-6)cos^2(2 * 45°)) - (3.085462×10^(-4) + 2.27×10^(-7)cos^2(45°)) * 0 + (7.254×10^(-11) + 1.0×10^(-18)cos^2(45°)) * (0)^2 - (1.517×10^(-17) + 6×10^(-20)cos^2(45°)) * (0)^3

Calculating the value gives: g(45°, 0) ≈ 9.80666553 m/s^2.

(d) 10,000 feet above mean sea level at the Antarctic Circle (≈66.5 degrees South latitude):

Latitude (φ) = -66.5°

Height (Z) = 10,000 feet = 3,048 meters

Substituting these values into the formula, we have:

g(-66.5°, 3,048) = 980.6160(1 - 0.0026373cos^2(-66.5°) + 5.9×10^(-6)cos^2(2 * -66.5°)) - (3.085462×10^(-4) + 2.27×10^(-7)cos^2(-66.5°)) * 3,048 + (7.254×10^(-11) + 1.0×10^(-18)cos^2(-66.5°)) * (3,048)^2 - (1.517×10^(-17) + 6×10^(-20)cos^2(-66.5°)) * (3,048)^3

Calculating the value gives: g(-66.5°, 3,048) ≈ 9.83065116 m/s^2.

(e) For a particular Z, the value of g obtained from this model is not the same at latitude φ as at latitude -φ. This is because the formula for g takes into account the latitude (φ) as a trigonometric term, specifically cos^2(φ). Since the cosine function is an even function (cos(-θ) = cos(θ)), the value of g will be different for φ and -φ, except when φ = 0 (equator) where cos(0) = cos(0) = 1.

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Answer:

air above a wild fire

Explanation:

the type of current formed in the pot is referred to as a convectional current which forms when the heated fluid rises making the cool fluid descend the same happens with the air above a wildfire a convectional current will form when the warm air rises and the cool air descends.the main reason for tge descending and ascending is that the air or liquid closest to the source of heat will get heated making it lighter but the air or liquid tgat was further from the source of heat will not be as light so as the warm fluid rises, the cool fluid descends.

Answer:

air above a wildfire

Explanation

The distance between charge 1 and 3 is equal to 20 cm and the force between charge 1 and 3 is equal to force between charge 1 and 3 that is 0.9 Newton and the force between charge 2 and 3 is equal to 1.41 Newton.

Electric charges are basic property of matter carried by some elementary particles that governs how the particles are affected by an electric or magnetic field. Electric charge, which can be positive or negative, occurs in discrete natural units and is neither created nor destroyed.

Coulomb's law states that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.

Given:

Magnitude on charge 1 = 4 μC

Magnitude on charge 2 = 4 μC

Magnitude on charge 3 = 1 μC

Distance between charge 1 and 2 = 32 cm

Distance of charge 3 from center of charge 1 and 2 = 12 cm

Distance between charge 1 and 3 can be find using Pythagoras theorem.

Distance between charge 1 and 3 = √(256 + 144)

Distance between charge 1 and 3 = 20cm

Similarly Distance between charge 2 and 3 = 20cm

Force between charge 1 and 3 can be found using coulomb law.

Force between charge 1 and 3 = (9 × 4 × 1)/40

Force between charge 1 and 3 = 36/40

Force between charge 1 and 3 = 0.9 Newton

Similarly using symmetry, force between charge 2 and 3 = 0.9 Newton

Force between charge 1 and 2 =(9 × 4 × 4)/102

Force between charge 1 and 2 = 1.41 newton

Therefore the distance between charge 1 and 3 is equal to 20 cm and the force between charge 1 and 3 is equal to force between charge 1 and 3 that is 0.9 Newton and the force between charge 2 and 3 is equal to 1.41 Newton.

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In the steady state, 5 × \(10^{18}\) electrons enter the wire every second 3.24 x \(10^{19}\) electrons enter wire B every second.

The electric current in a wire is given by:

I = Anev

Where I is current, A is the cross-sectional area of the wire, n is the number of mobile electrons per unit volume, e is the charge of an electron, and v is the drift velocity of the electrons.

Let's start by finding the current in wire A:

\(I_A\) = \(A_A\) × \(n_A\) × e × \(v_A\)

where \(A_A\) is the cross-sectional area of wire A, \(n_A\) is the number of mobile electrons per unit volume in wire A, and \(v_A\) is the drift velocity of electrons in wire A.

We are given that 3e18 electrons enter wire A every second, so the current in wire A is:

\(I_A\) = 3e18 × e

Now let's find the current in wire B:

I_B = \(A_B\) × \(n_B\) × e × \(v_B\)

where \(A_B\) is the cross-sectional area of wire B, \(n_B\) is the number of mobile electrons per unit volume in wire B, and \(v_B\) is the drift velocity of electrons in wire B.

We are told that wire B has 4 times the cross-sectional area of wire A, 1.7 times as many mobile electrons per cubic centimeter, and the same mobility as wire A. Therefore:

\(A_B\) = 4 × \(A_A\)

\(n_B\) = 1.7 × \(n_A\)

\(v_B\) = \(v_A\)

Substituting these values into the equation for current in wire B, we get:

\(I_B\) = 4 × \(A_A\) × 1.7 × \(n_A\) × e × \(v_A\)

Simplifying, we get:

\(I_B\) = 10.8 × \(I_A\)

So the number of electrons entering wire B every second is:

\(I_B\) / e = 10.8 × \(I_A\) / e = 10.8 × 3e18 = 3.24e19 electrons/second.

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The spring constant is:

                                             \(\Large\displaystyle\text{$\begin{aligned}k &= 612.5\ \dfrac{\text{N}}{\text{m}}\end{aligned}$}\)

To calculate the spring constant we must remember the law for it, the Hooke's Law:

                                            \(\Large\displaystyle\text{$\begin{aligned}\vec{F} = -k\Delta \vec{x} \end{aligned}$}\)

Where k is the spring constant [N/m].

So if the mass is suspended it means that its weight is equal to the elastic force (values), then we can write:

                                           \(\Large\displaystyle\text{$\begin{aligned}\vec{W} = k\Delta \vec{x} \end{aligned}$}\)

Therefore:

                                          \(\Large\displaystyle\text{$\begin{aligned}mg &= k\Delta x \\ \\k &= \dfrac{mg}{\Delta x} \\ \\\end{aligned}$}\)

Now we just have to put the values and calculate:

                                              \(\Large\displaystyle\text{$\begin{aligned}k &= \dfrac{mg}{\Delta x} \\ \\k &= \dfrac{5\cdot 9.8}{0.08} \\ \\k &= 612.5\ \dfrac{\text{N}}{\text{m}}\end{aligned}$}\)

I hope you liked it

Any doubt? write in the comments and I'll help you

Answer:

Below

Explanation:

Potential Energy of the cars is turned into Kinetic Energy (ignoring friction)

mgh = 1/2 mv^2       divide both sides by 'm'

gh = 1/2 v^2               solve for v

v = sqrt (2gh)            Note that the mass of the cars is irrelevent

For the first car :     25 cm = .25 m

                                  v = sqrt (2 * 9.81* .25) = 2.2 m/s

 Similarly , for car 2      v = 4.8 m/s           (115 cm = 1.15 m)

The vertical component of initial velocity must be 3.33 m/s upwards to barely clear the 1.00 m high net and the ball will hit the ground 27.06 m beyond the net.

Given: Initial horizontal velocity (Vox) = 24.5 m/s, Initial vertical velocity (Voy) = ?Initial elevation (h) = 1.43 mDistance from the net (x) = 10.4 m, Net height (y) = 1.00 m(a) What vertical component of initial velocity must be given to the ball, such that it barely clears the 1.00 m high net?We can use the kinematic equation, `y = Voy*t + (1/2)*a*t^2 + h`Here, `a = -9.8 m/s^2` (acceleration due to gravity acting downwards)At the highest point of the trajectory (i.e. when the ball barely clears the net), the vertical component of velocity becomes zero.

So, `Vf = 0 m/s`. Now, let's calculate the time taken for the ball to reach the highest point.`Vf = Vo + a*t` `0 = Voy + (-9.8)*t` `t = Voy/9.8`We can use this value of time in the above kinematic equation and put `Vf = 0` to get the initial vertical velocity. `y = Voy*t + (1/2)*a*t^2 + h``0 = Voy*(Voy/9.8) + (1/2)*(-9.8)*(Voy/9.8)^2 + 1.43``0 = Voy^2/9.8 - Voy^2/19.6 + 1.43``0.7167 = Voy^2/19.6`So, `Voy = sqrt(0.7167*19.6)` `= 3.33 m/s` (upwards)

Therefore, the vertical component of initial velocity must be 3.33 m/s upwards to barely clear the 1.00 m high net.

(b) How far beyond the net will the ball hit the ground? The time taken for the ball to reach the highest point is `t = Voy/9.8` `= 3.33/9.8` `= 0.34 s`

Therefore, the total time of flight of the ball is `2*t = 0.68 s`. Horizontal distance travelled by the ball in `0.68 s` is given by: `x = Vox*t``x = 24.5*0.68``x = 16.66 m` (approx). So, the ball will hit the ground `10.4 + 16.66 = 27.06 m` from the base line (i.e. beyond the net).

Therefore, the ball will hit the ground 27.06 m beyond the net.

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Answer:

\(20.61\times 10^{5}Nm^{-2}\)

Explanation:

We are given that

h=20 cm=0.20m

1m=100 cm

Density of water=\(\rho=1.0\times 10^3 kg/m^3\)

We have to find the pressure experienced by driver at that depth.

Atmospheric pressure, P0=101 kPa

Pressure experience by driver

\(P=\rho hg+P_0\)

Where \(g=9.8\)

\(P=1.0\times 10^3\times 0.20\times 9.8+101\)

\(P=2061KPa\)

1 KPa=\(1000Nm^{-2}\)

\(P=2061\times 1000=2061000N/m^2\)

\(P=20.61\times 10^{5}Nm^{-2}\)

The same-side exterior angles are not congruent, they are supplementary. The same side exterior angles are formed and they have a sum of 180 degrees.

Congruent refers to having the same precise size and shape. Even if we flip, turn, or rotate the forms, their shape and size need to remain constant. If it is possible to superimpose one geometric figure onto the other such that they correspond throughout, then the two are said to be congruent, or to be in the relation of congruence.

In general relativity, a congruence (more precisely, a congruence of curves) is the collection of integral curves of a (never vanishing) vector field in a four-dimensional Lorentzian manifold that is used to physically represent spacetime.

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The imaging properties of lenses and mirrors discussed in this write-up are based on the paraxial ray approximation,  which stipulates that the light rays involved make only small angles with the optic axis.

The mirror is a glass with one side coated with silver lining which help to  produces an image by reflection on only one surface. Whereas , the lens is a transparent substance that produces images by refraction

A paraxial ray is a ray that makes a small angle towards the optical axis of the system . When light rays make small angles with the optic axis, it is known as paraxial ray approximation.

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