Ask question

Ask question

All categories

2 years ago

12

A beam of unpolarized light shines on a stack of five ideal polarizers, set up so that the angles between the polarization axes

of pairs of adjacent polarizers are all equal. The intensity of the transmitted beam is reduced from the intensity of the initial beam by a factor of phi=0.277 . Find the angle theta between the axes of each pair of adjacent polarizers.

1 answer:

ddd [48]2 years ago

7 0

Answer:

$21.75$

Explanation:

$n$ = number of polarizers through which light pass through = 5

$\theta$ = Angle between each pair of adjacent polarizers

$I_{o}$ = Intensity of unpolarized light

$I_{n}$ = Intensity of transmitted beam after passing all polarizers

It is given that

$\frac{I_{n}}{I_{o}}= 0.277$

we know that the intensity of light after passing through "n" polarizers is given as

$I_{n} = (0.5) I_{o} Cos^{2n-2}\theta$

$\frac{I_{n}}{I_{o}} = (0.5) Cos^{2n-2}\theta$

inserting the values

$0.277 = (0.5)Cos^{2(5)-2}\theta$

$0.554 = Cos^{8}\theta$

$Cos\theta = 0.9288$

$\theta = Cos^{-1}(0.9288)$

$\theta = 21.75$

You might be interested in

A frog leaps up from the ground and lands on a step 0.1 m above the ground 2 s later. We want to find the

mash [69]

Answer:

$\Delta x = v_0 t + \frac{1}{2}at^2$

Explanation:

To solve this problem, we can use the following suvat equation:

$\Delta x = v_0 t + \frac{1}{2}at^2$

where

$\Delta x$ is the vertical displacement of the frog

$v_0$ is the initial vertical velocity

t is the time

a is the acceleration

We have chosen this formula because apart from $v_0$ , all the other quantities are known. In fact:

$\Delta x =0.1 m$ is the vertical displacement

t = 2 s is the total time of flight

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (negative because it is downward)

Therefore, solving for $v_0$ , we find the initial velocity of the frog:

$v_0 = \frac{\Delta x-\frac{1}{2}at^2}{t}=\frac{0.1-\frac{1}{2}(-9.8)(2)^2}{2}=9.85 m/s$

4 0

2 years ago

In a house the temperature at the surface of a window is 28.9 °C. The temperature outside at the window surface is 7.89 °C. Heat

Alenkasestr [34]

Answer:

-13.18°C

Explanation:

To develop the problem it is necessary to consider the concepts related to the thermal conduction rate.

Its definition is given by the function

$\frac{Q}{t} = \frac{kA\Delta T}{d}$

Where,

Q = The amount of heat transferred

t = time

k = Thermal conductivity constant

A = Cross-sectional area

$\Delta T =$ The difference in temperature between one side of the material and the other

d= thickness of the material

The problem says that there is a loss of heat twice that of the initial state, that is

$Q_2 = 2*Q_1$

Replacing,

$kA\frac{\Delta T_m}{x} = 2*kA\frac{\Delta T}{x}$

$\frac{\Delta T}{x}=2*\frac{\Delta T}{x}$

$\frac{T_i-T_o}{x} = 2\frac{T_1-T_2}{x}$

$\frac{28.9-T_o}{x} = 2\frac{28.9-7.86}{x}$

Solvinf for $T_o$ ,

$T_o = -13.18$

Therefore the temprature at the outside windows furface when the heat lost per second doubles is -13.18°C

3 0

2 years ago

Aaron Agin nodded off while driving home from play practice this past Sunday evening. His 1500-kg car hit a series of guardrails

Inessa [10]

Answer: 6.48m/s

Explanation:

First, we know that Impulse = change in momentum

Initial velocity, u = 19.8m/s

Let,

Velocity after first collision = x m/s

Velocity after second collision = y m/s

Also, we know that

Impulse = m(v - u). But then, the question said, the guard rail delivered a "resistive" impulse. Thus, our impulse would be m(u - v).

5700 = 1500(19.8 - x)

5700 = 29700 - 1500x

1500x = 29700 - 5700

1500x = 24000

x = 24000/1500

x = 16m/s

Also, at the second guard rail. impulse = ft, so that

Impulse = 79000 * 0.12

Impulse = 9480

This makes us have

Impulse = m(x - y)

9480 = 1500(16 -y)

9480 = 24000 - 1500y

1500y = 24000 - 9480

1500y = 14520

y = 14520 / 1500

y = 9.68

Then, the velocity decreases by 3.2, so that the final velocity of the car is

9.68 - 3.2 = 6.48m/s

5 0

1 year ago

A cylindrical wire has a resistance R and resistivity ρ. If its length and diameter are BOTH cut in half, what will be its resis

snow_lady [41]

Answer:

The resistance will be 2×R

Explanation:

We note that the resistivity of a cylindrical wire is given by the following relation;

$\rho = \frac{RA}{L}$

Where:

ρ = Resistivity of the wire

R = The wire resistance

A = Cross sectional area of the wire = π·D²/4

L = Length of the wire

Rearranging, we have;

$R= \frac{\rho L}{A}$

If the length and the diameter are both cut in half, we have;

L₂ = L/2

A₂ =π·D₂²/4 = $\pi \cdot \left (\frac{D}{2} \right )^{2} \times \frac{1}{4} = \pi \cdot \frac{D^{2}}{16} = A/4$

Therefore, the new resistance, R₂ can be expressed as follows;

$R_2= \frac{\rho \frac{L}{2} }{\frac{A}{4} } = \rho \frac{L}{2} \times \frac{4}{A} = 2 \times \frac{\rho L}{A}$

Hence, the new resistance R₂ = 2×R, that is the resistance will be doubled.

8 0

2 years ago

the grid in a triode is kept negatively charged to prevent… a. the variations in voltage from getting too large. b. electrons be

Nezavi [6.7K]

D:the electrons from being attracted to the grid instead of the anode

5 0

1 year ago

Read 2 more answers