For a certain optical medium the speed of light varies from a low value of 1.80 × 10 8 m/s for violet light to a high value of 1.92 × 10 8 m/s for red light. Calculate the range of the index of refraction n of the material for visible light.

Answers

Answer 1

Answer:

1.56 - 1.67

Explanation:

Refractive index of any material is given as the ratio of the speed of light in a vacuum to the speed of light in that medium.

Mathematically, it is given as:

n = c/v

Where c is the speed of light in a vacuum and v is the speed of light in the medium.

Given that the speed of light in the optical medium varies from 1.8 * 10^8 m/s to 1.92 * 10^8 m/s, we can find the range of the refractive index.

When the speed is 1.8 * 10^8 m/s, the refractive index is:

n = (3 * 10^8) / (1.8 * 10^8)

n = 1.67

When the speed is 1.92 * 10^8 m/s, the refractive index is:

n = (3 * 10^8) / (1.92 * 10^8)

n = 1.56

Therefore, the range of values of the refractive index of the optical medium is 1.56 - 1.67.

Related Questions

A 2,000 kg car travels with a tangentialvelocity of 12 m/s around a circular track with a radius of 30 meters. What is the car's rate of centripetal acceleration?
HELP ASAP PLEASE!!!In which direction(s) does the ground shake during an earthquake?A. sidewaysB. up and downC. back and forthD. all of the above
If 2050 J of heat are added to a 150 g object its temperature increases by 15°C.(a) What is the heat capacity of this object? (b) What is the object's specific heat?
A horizontal cylindrical tank 8.00 ft in diameter is half full of oil (60.0 Ib/ft3). Find the force on one end
If you drew magnetic field lines for this bar magnet, which statement would be true

A physics cart has a projectile launcher mounted on top. While traveling on a straight track at 0.500 m/s, a projectile is fired. It lands back in the same place on top of the launcher after the cart has moved a distance of 2.30 m. In the frame of reference of the cart, (a) at what angle was the projectile fired and (b) what was the initial velocity of the projectile? (c) What is the shape of the projectile as seen by an observer on the cart? A physics student is watching the demonstration from a classroom seat. According to the student, (d) what is the shape of the projectile’s path, and (e) what is its initial velocity?

Answers

Answer:

(a) $90^(\circ)$

(b) Initial velocity of the projectile is 22.54 m/s

(d) Parabolic

(e) The initial velocity is 23.04 m/s

Solution:

As per the question:

Velocity of the cart, v = 0.500 m/s

Distance moved by the cart, d = 2.30 m

Now,

(a) The projectile must be fired at an angle of $90^(\circ)$ so that it mounts on the top of the cart moving with constant velocity.

(b) Now, for initial velocity, u':

Time of flight is given by;

$T = (D)/(v)$ (1)

where

T = Flight time

D = Distance covered

(b) The component of velocity w.r.t an observer:

Horizontal component, $v_(x) = u'cos\theta$

Vertical component, $v_(y) = u'sin\theta - gT$

Also, the vertical component of velocity at maximum height is zero, $v_(y) = 0$

Therefore, $T = (u')/(g)$

Total flight time, $(2u')/(g)$ (2)

Now, from eqn (1) and (2):

$u' = (gD)/(2v)$

$u' = (9.8* 2.30)/(2* 0.500) = 22.54 m/s$

(d) The shape of the path of the projectile seen by the physics student outside the reference frame of the cart is parabolic

(e) The initial velocity is given by:

u = u' + v = 22.54 + 0.5 = 23.04 m/s

Box 1 and box 2 are whirling around a shaft with a constant angular velocity of magnitude ω. Box 1 is at a distance d from the central axis, and box 2 is at a distance 2d from the axis. You may ignore the mass of the strings and neglect the effect of gravity. Express your answer in terms of d, ω, m1 and m2, the masses of box 1 and 2. (a) Calculate TB, the tension in string B (the string connecting box 1 and box 2). (b) Calculate TA, the tension in string A (the string connecting box 1 and the shaft).

Answers

Answer:

a) TB = m2 * w^2 * 2*d

b) TA = m1 * w^2 * d + m2 * w^2 * 2*d

Explanation:

The tension on the strings will be equal to the centripetal force acting on the boxes.

The centripetal force is related to the centripetal acceleration:

f = m * a

The centripetal acceleration is related to the radius of rotation and the tangential speed:

a = v^2 / d

f = m * v^2 / d

The tangential speed is:

v = w * d

Then

f = m * w^2 * d

For the string connecting boxes 1 and 2:

TB = m2 * w^2 * 2*d

For the string connecting box 1 to the shaft

TA = m1 * w^2 * d + m2 * w^2 * 2*d

. An electron moving at 4.00×103m/s in a 1.25-T magnetic field experiences a magnetic force of 1.40×10−16N . What angle does the velocity of the electron make with the magnetic field? There are two answers.

Answers

Answer: 9.59° and 350.41°

Explanation: The formulae that relates the force F exerted on a moving charge q with velocity v in a magnetic field of strength B is given as

F =qvB sin x

Where x is the angle between the strength of magnetic field and velocity of the charge.

q = 1.609×10^-19 C

v = 4×10³ m/s

B = 1.25 T

F = 1.40×10^-16 N

By substituting the parameters, we have that

1.40×10^-16 = 1.609×10^-19 × 4×10³ × 1.25 × sinx

sin x = 1.40×10^-16/ 1.609×10^-19 × 4×10³ × 1.25

sin x = 1.40×10^-16 /8.045*10^(-16)

sin x = 0.1666

x = 9.59°

The value of sin x is positive in first and fourth quadrant.

Hence to get the second value of x, we move to the 4th quadrant of the trigonometric quadrant which is 360 - x

Hence = 360 - 9.59 = 350.41°

A train station bell gives off a fundamental tone of 505 Hz as the train approaches the station at a speed of 27.6 m/s. If the speed of sound in air on that day is 339 m/s, what will be the apparent frequency of the bell to an observer riding the train

Answers

Answer:

Apparent frequency of the bell to the observer is 546.12 Hz

Explanation:

The frequency of train bell (frequency of source) = 505 Hz

The speed of train (observer) = 27.6 m/s

The speed of sound in the air is (velocity of sound) = 339 m/s

The apparent frequency of the bell to the observer is calculated as follows:

Apparent frequency of bell to the observer.

$= \text{frequency of source} * (Observer + velocity \ of \ sound )/( velocity \ of \ sound ) \n= 505 * (27.6 + 339)/(339) \n= 546.12 Hz$

a mass of .4 kg is raised by a vertical distance of .450 m in the earth's gravitational field. what is the change in its gravitational potential energy

Answers

Answer:

E = 1.76 J

Explanation:

Given that,

Mass of an object, m = 0.4 kg

It moves by a vertical distance of 0.45 m in the Earth's gravitational field.

We need to find the change in its gravitational potential energy. It can be given by the formula as follow :

$E=mgh\n\nE=0.4* 9.8* 0.45\n\nE=1.76\ J$

So, the change in its gravitational potential energy is 1.76 J.

The equations for single-slit and multiple-slit interference both contain the variable θ. For the multiple-slit case, the angle is: a. the angular location of the first order minimum in the diffraction pattern. Which means at this point the light experiences constructive interference.
b. the angular location of the first order minimum in the diffraction pattern. Which means at this point the light experiences destructive interference.
c. the angular location of bright interference maxima in the pattern. Which means at this point the light experiences constructive interference.
d. the angular location of bright interference maxima in the pattern. Which means at this point the light experiences destructive interference.

Answers

Answer:

the answers the correct one is c

Explanation:

The diffraction pattern for a slit is

a sin θ = m λ

Where a is the width of the slit, λ the wavelength, m the order of destructive interference and θ the angle where the interference occurs.

The expression for multi-slit diffraction (diffraction grating) is

d sin θ = m λ

Where d is the distance between slits, λ the wavelength m the order of the diffraction maximums and θ the angle for these maximums.

When we compare the expressions of the answers the correct one is c

Other Questions

Which statement about thin lenses is correct? In each case, we are considering only a single lens. A. A diverging lens always produces a virtual inverted image. B. A converging lens always produces a real inverted image. C. A converging lens sometimes produces a real erect image. D. A diverging lens produces a virtual erect image only if the object is located within the focal point of the lens. E. A diverging lens always produces a virtual erect image.

During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts. The limb exerts a downward force of 176 N on the wire. The left section of the wire makes an angle of 12.5° relative to the horizontal and sustains a tension of 413 N. Find the (a) magnitude and (b) direction (as an angle relative to horizontal) of the tension that the right section of the wire sustains.

A rabbit is moving in the negative x-direction at 1.10 m/s when it spots a predator and accelerates to a velocity of 13.9 m/s along the negative y-axis, all in 1.20 s. Determine the x-component and the y-component of the rabbit's acceleration. (Enter your answers in m/s2. Indicate the direction with the signs of your answers.) HINT

Which state of matter is most similar to solids