Unpolarized light is passed through an optical filter that is oriented in the vertical direction. 1) If the incident intensity of the light is 90 W/m2, what is the intensity of the light that emerges from the filter? (Express your answer to two significant figures.)

Answers

Answer 1
Answer:

Answer:

45 W/m^2

Explanation:

Intensity of light, Io = 90 W/m^2

According to the law of Malus

I=I_(0)Cos^(2)\theta

The average value of Cos^θ is half

So, I = Io/2

I = 90 /2

I = 45 W/m^2

Answer 2
Answer:

Final answer:

Unpolarized light, when passed through a polarizer, reduces its intensity by half. So, the intensity if the light that emerges from a vertical filter will be 45 W/m².

Explanation:

Given that the incident intensity of the unpolarized light is 90 W/m², when passed through a vertically oriented optical filter, the emerging light will be polarized and will have its intensity halved as it's the property of a polarizing filter to decrease the intensity of unpolarized light by a factor of 2. The formula used in this process is I = Io cos² θ. In the case of unpolarized light passing through a single polarizer, θ is 0. So, the formula simplifies to I = Io/2.

Therefore, the intensity of the light that emerges from the vertically oriented optical filter is: I = 90 W/m² / 2 = 45 W/m².

Learn more about Polarization of Light here:

brainly.com/question/29217577

#SPJ3


Related Questions

The position of a particle is given by the function x=(5t3−8t2+12)m, where t is in s. at what time does the particle reach its minimum velocity?
A 100 A current circulates around a 2.00-mm-diameter superconducting ring. A. What is the ring's magnetic dipole moment?B. What is the on-axis magnetic field strength 4.70 cm from the ring?
A string is attached to the rear-view mirror of a car. A ball is hanging at the other end of the string. The car is driving around in a circle, at a constant speed. Which of the following lists gives all of the forces directly acting on the ball?a. tensionb. tension and gravityc. tension, gravity, and the centripetal forced. tension, gravity, the centripetal force, and friction
A particle (q = 5.0 nC, m = 3.0 μg) moves in a region where the magnetic field has components Bx = 2.0 mT, By = 3.0 mT, and Bz = −4.0 mT. At an instant when the speed of the particle is 5.0 km/s and the direction of its velocity is 120° relative to the magnetic field, what is the magnitude of the acceleration of the particle in m/s2?
5. The order is to give 600 mg of Ampicillin IM q8h. The directions for dilution on the 2 gm vial reads: Reconstitute with 4.8 mL of sterile water to obtain a concentration of 400 mg per mL. How many mL will you administer per dose?

Fuel cells have been developed that can generate a large amount of energy. For example, a hydrogen fuel cell works by combining hydrogen and oxygen gas to produce water and electrical energy. If a fuel cell can generate 10.0 kilowatts of power and the current is 15.8 amps, what is the voltage of the electricity?A.
0.63 volts
B.
158volts
C.
633 volts
D.
158,000 volts
E.
5.8 volts

Answers

The voltage of the electricity will be 632.9 V. Electric power is found as the multiplication of the voltage and current. Option B is correct.

What is electric power?

Electric power is the product of the voltage and current. Its unit is the watt. It is the rate of the electric work done.

The given data in the problem is;

V is the voltage = ? Volt (V)

Electric current (I)= 15.8 amps (A)

P is the power =10.0 kilowatts =10⁴ watt

The formula for the power is given as;

\rm P= V I \n\n\ 10^4= V * 15.8 \n\n V=632.9 \ V

The voltage of the electricity will be 63.29 V.

Hence, option B is correct.

To learn more about the electric power, refer to the link;

brainly.com/question/12316834

#SPJ2

Hmmm. Kilowatts should be converted to watts. Simply just move the decimal place to the right three times.

10,000 W / 15.8 A = V

632.9, or 633.

determine exactly where to place a cart on the track so that it rolls down the track, flies through the air, and lands precisely at 1) the green line, 2) the red line, and 3) the blue line, on the first try.

Answers

Answer: i think you should place it on the red line

Explanation:

hope this helps

and need brainliest

A concave mirror is to form an image of the filament of a headlight lamp on a screen 8.50 m from the mirror. The filament is 6.00 mm tall, and the image is to be 37.5 cm tall. Part A: How far in front of the vertex of the mirror should the filament be placed?Part B: to what radius of curvature should you grind the mirror?

Answers

Answer:13.6 cm

Explanation:

Given

v(image distance)=-8.5 m

height of object(h_1)=6 mm

height of image (h_2)=37.5 cm

and magnification of concave mirror is given by m=(-v)/(u)=(-h_2)/(h_1)

m=(-37.5* 10)/(6)=-62.5

-62.5=(8.5* 100)/(u)

u=13.6 cm

so object is at a distance of 13.6 cm from mirror.

for focal length

(1)/(f)=(1)/(v)+(1)/(u)

(1)/(f)=(-1)/(850)+(-1)/(13.6)

(1)/(f)=-0.00117-0.0735

f=-13.4 cm

thus radius of curvature of mirror is R=2f=26.8 cm

Final answer:

The filament of the headlight lamp should be placed about 0.85 m in front of the vertex of the mirror. The radius of curvature for the concave mirror should be approximately 0.85 m.

Explanation:

To determine how far in front of the vertex of the mirror the filament should be placed, we can use the mirror equation:

1/f = 1/do + 1/di

Where f is the focal length of the concave mirror, do is the object distance, and di is the image distance.

With the given information, we have:

do = ?

di = 8.50 m

Using the magnification formula:

magnification = -di/do

By substituting the values we know, we can solve for do:

37.5 cm / 6.00 mm = -8.50 m / do

Solving for do, we find that do ≈ - 0.85 m.

Since the object distance cannot be negative, we conclude that the filament of the headlight lamp should be placed about 0.85 m in front of the vertex of the mirror.

To find the radius of curvature for the concave mirror, we use the mirror formula:

1/f = 1/do + 1/di

With do = -0.85 m and di = 8.50 m, we can rearrange the formula to solve for f:

1/f = 1/-0.85 + 1/8.50

1/f ≈ -1.1765

Solving for f, we find that the focal length is approximately 0.85 m.

Learn more about Concave mirror here:

brainly.com/question/3555871

#SPJ3

A smart phone charger delivers charge to the phone, in the form of electrons, at a rate of -0.75. How many electrons are delivered to the phone during 27 min of charging?

Answers

Answer:

The no. of electrons is 7.59* 10^(21)

Solution:

According to the question:

The rate at which the charge is delivered is given by:

(dQ)/(dt) = - 0.75

Now,

\int_(0)^(Q)dQ = - 0.75\int_(0)^(27 min) dt

Q = -0.75t|_(0)^(27 min)

Q= -0.75* 27* 60 = - 1215 C

No. of electrons, n can be calculated from the following relation:

Q = ne

where

e = electronic charge =1.6* 10^(- 19) C

Thus

n = (Q)/(e)

n= (1215)/(1.6* 10^(- 19))

n = 7.59* 10^(21)

Which statementsabout a neutral atom are correct? Check all that apply.1. A neutral atom is composed of bothpositively and negatively charged particles.
2. Positively charged protons are located in the tiny, massivenucleus.
3. The positively chargedparticles in the nucleus are positrons.
4. The negatively chargedelectrons are spread out in a "cloud" around thenucleus.
5. The electrons areattracted to the positively charged nucleus.
6. The radius of the electroncloud is twice as large as the radius of the nucleus.

Answers

Answer:

1, 2, 4, 5 are correct

Explanation:

1) This is true because In a neutral atom, the number of positive charges (protons) is equal to the number of negative charges (electrons).

2) This is true because the mass of the atom which is made up of the protons and neutrons, is located in the tiny nucleus.

3) This is not true because the positively charged particles in the nucleus are called protons.

4) This is true because electrons move around the nucleus in diffuse areas known as orbitals.

5) This is true because opposite charges attract each other. And electron is a negative charge.

6) This is not true because the radius of the electron cloud is normally 10,000 times larger than the radius of the nucleus.

How much electrical energy is used by a 400 W toaster that is operating for 5minutes?
A. 2000 J
B. 75,000 J
C. 120,000 J
D. 300,000 J

Answers

The electrical energy used by a 400 W toaster that is operating for 5 minutes will be 120,000 J.Option C is correct.

What is the power output?

The rate of the work done is called the power output. It is denoted by P.Its unit of a watt. It is the ratio of the work done or the enrgy to the time period.

The given data in the problem is;

E is the electrical energy

P is the power output =  400 W

t is the time period = 5 minutes

The power output is given as;

\rm P= (E)/(t) \n\n\ E= P * t \n\n\ E= 400 * 300 \n\n\ E=120,000 \ J

Hence the electrical energy used by a 400 W toaster that is operating for 5 minutes will be 120,000 J.Option C is correct.

To learn more about the power output refer to the link;

brainly.com/question/22285866

Answer:

The answer is C. 120,000 J.

Explanation: