Polaroid Vision in a Spider Experiments show that the ground spider Drassodes cu

Question

Polaroid Vision in a Spider Experiments show that the ground spider Drassodes cupreus uses one of its several pairs of eyes as a polarization detector. In fact, the two eyes in this pair have polarization directions that are at right angles to one another. Suppose linearly polarized light with an intensity of 800 W/m2 shines from the sky onto the spider, and that the intensity transmitted by one of the polarizing eyes is 242 W/m2 .

Part A

For this eye, what is the angle between the polarization direction of the eye and the polarization direction of the incident light?

Part B

What is the intensity transmitted by the other polarizing eye?

Explanation / Answer

Here ,

input intensity , Io = 800 W/m^2

Iout = 242 W/m^2

part A)

angle between the polarization direction of the eye and the polarization direction of the incident light is theta

Iout = Io * cos^2(theta)

242 = 800 * cos^2(theta)

solving for theta

theta = 56.6 degree

the angle between the polarization direction of the eye and the polarization direction of the incident light is 56.6 degree

part B)

Now, as the angle between the eyes is 90 degree

intensity transmitted by the other polarizing eye = Iin* cos^2(90 -theta)

intensity transmitted by the other polarizing eye = 800 * cos^2(90 - 56.6)

Here ,

input intensity , Io = 800 W/m^2

Iout = 242 W/m^2

part A)

angle between the polarization direction of the eye and the polarization direction of the incident light is theta

Iout = Io * cos^2(theta)

242 = 800 * cos^2(theta)

solving for theta

theta = 56.6 degree

the angle between the polarization direction of the eye and the polarization direction of the incident light is 56.6 degree

part B)

Now, as the angle between the eyes is 90 degree

intensity transmitted by the other polarizing eye = Iin* cos^2(90)

intensity transmitted by the other polarizing eye =

the Here ,

input intensity , Io = 800 W/m^2

Iout = 242 W/m^2

part A)

angle between the polarization direction of the eye and the polarization direction of the incident light is theta

Iout = Io * cos^2(theta)

242 = 800 * cos^2(theta)

solving for theta

theta = 56.6 degree

the angle between the polarization direction of the eye and the polarization direction of the incident light is 56.6 degree

part B)

Now, as the angle between the eyes is 90 degree

intensity transmitted by the other polarizing eye = Iin* cos^2(90 - 56.6)

the intensity transmitted by the other polarizing eye = 800 * cos^2(33.4)

the intensity transmitted by the other polarizing eye = 557.6 W/m^2

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