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zmey [24]
3 years ago
14

If you have two linear polarizers whose extinction axis are 90 degrees relative to eachother, no light passes through the system

.
Where can you put a third polarizer to allow light to pass through the system and what angle should it be at in order to allow the maximum amount of light to be transmitted?
Physics
1 answer:
marin [14]3 years ago
8 0

Answer:

The third polarizer can be placed midway between the first two polarizers with its extinction axis at 45° from either polarizer to maximize the amount of light that is transmitted (one-eight).

Explanation:

If light is incident on a polarizer, it allows only light that is parallel to its 'pass-through' axis to pass through untouched.

Light whose electric direction/vector is perpendicular to the 'pass through' axis will not pass through at all. Light whose electric direction/vector points in other directions (apart from those whose direction is parallel or perpendicular) passes through according to the magnitude of the component that is parallel to the 'pass-through' axis.

The polarizer blocks half of the incident light rays and the transmitted light is polarized in the direction of the 'pass-through' axis.

A new polarizer now place at a distance from the first polarizer with its 'pass-through' axis perpendicular to the first polarizer cancel out all the light that comes through from the first polarizer. Since the light electric vector needs to be parallel to the axis of the polarizer to pass through and all the parallelized light from the first polarizer are now incident perpendicularly to the axis of the second polarizer, no light rays pass through.

But, a third polarizer can be placed midway between the first two polarizers with its axis positioned at 45° from either polarizer. Thereby allowing exactly half of the light from the first polarizer to pass through. The explanation is just like that for the first one. (Light whose electric direction/vector points in other directions (apart from those whose direction is parallel or perpendicular) passes through according to the magnitude of the component that is parallel to the 'pass-through' axis).

Then the resultant from the middle polarizer reaches the initial second polarizer and half of the light is let through again. So that, at the end of the day, (1/2) × (1/2) × (1/2) of the initial incident ray is let through.

That is, to maximize the amount of light that is transmitted (one-eight of initial incident ray) the third Polaroid is place midway between the first two and at angle 45° to either one.

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Explanation:

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3 years ago
A toaster using a Nichrome heating element operates on 120 V. When it is switched on at 28 ∘С, the heating element carries an in
sukhopar [10]

Answer:

The final temperature of the element = 262.67°C

The power dissipated in the heating element initially = 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W

Explanation:

Our given parameters include;

A Nichrome heating element operates on 120 V.

Voltage (V) = 120V

Initial Current (I₁) = 1.36 A

Initial Temperature (T₁) = 28°C

Final Current (I₂) = 1.23 A

Final Temperature (T₂) = unknown ????

Temperature dependencies of resistance is given by:

R_{T(2)}=R_1[1+\alpha (T_2-T_1)]            ----------------------    (1)

in which R₁ is the resistance at temperature T₁

R_{T(2) is the resistance at temperature T₂

Given that V= IR

R = \frac{V}{I}

Therefore, the resistance at temperature 28°C is;

R_{28}= \frac{120V}{1.36A}

= 88.24Ω

R_{T(2) = \frac{120V}{1.23A}

= 97.56Ω

From (1) above;

R_{T(2)}=R_1[1+\alpha (T_2-T_1)]      

97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]

\frac{97.56}{88.24}= 1+(4.5*10^{-4})(T-28^0C)

1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)

0.1056 = 4.5×10⁻⁴(T₂-28°C)

\frac{0.1056}{4.5*10^{-4}}= T-28^0C

T - 28° C = 234.67

T = 234.67 + 28° C

T = 262.67 ° C

(b)

What is the power dissipated in the heating element initially and when the current reaches 1.23 A

The power dissipated in the heating element initially can be calculated as:

P = I²₁R₂₈

P = (1.36A)²(88.24Ω)

P = 163.209 W

P ≅ 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:

P= I^2_2R_{T^0C

P = (1.23)²(97.56Ω)

P = 147.598524

P ≅ 147.60 W

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