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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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Answer:

a. Wavelength = λ = 20 cm

b. Next distance of maximum intensity will be 40 cm

Explanation:

a. The distance between the two speakers is 20cm. SInce the intensity is maximum which refers that we have constructive interference and the phase difference must be an even multiple of π and equivalent path difference is nλ.

Now when distance increases upto 30 cm between the speakers, the sound intensity becomes zero which means that there is destructive interference and equivalent path is now increased from nλ to nλ + λ/2.

This we get the equation:

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b. at what distance, sound intensity will be maximum again.

For next point calculation for maximum sound intensity, the path difference must be increased (n+1) λ. The distance must increase by λ/2 from the point of zero intensity.

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