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Svetllana [295]
3 years ago
10

The gnaphosid spider Drassodes cupreus has evolved a pair of lensless eyes for detecting polarized light. Each eye is sensitive

to polarized light in perpendicular directions. Near sunset, the spider leaves its nest in search of prey. Light from overhead is linearly polarized and indicates the direction the spider is moving. After the hunt, the spider uses the polarized light to return to its nest. Suppose that the spider orients its head so that one of these two eyes detects light of intensity 700 W/m2and the other eye detects zero light intensity.
Part A: What intensity does the first eye (intensity 700 W/m2 ) detect if the spider now rotates its head 20 ∘ from the previous orientation?

Part B:What intensity does the first eye (zero intensity) detect if the spider now rotates its head 20 ∘ from the previous orientation?
Physics
1 answer:
BartSMP [9]3 years ago
5 0

Answer:

Part A

The intensity is  I = 618 W/m^2  

Part B

The intensity is  I_1= 81.884 W/m^2

Explanation:

From the question we are told that

       The intensity of the light detected by first eye is I = 700 W/m^2

Now at initial state according the question the light  ray is perpendicular to the eye so it means that it is at 90° the eye

Now the first question is to obtain the intensity the first eye (the first in this case is the one focused on the light  )would detect when the head is rotated by 20°  its previous orientation

This  is mathematically evaluated  as

                   I = I_i cos^2 ( 20^o)

                    I = 700\  cos^2 (20)

                    I = 618 W/m^2  

Now the second  question is to obtain the intensity the first eye (the first eye  in this case is the one that is not  focused on  the light  )would detect when the head is rotated by 20°  its previous orientation

Now in this case the angle between the eye and the light is 90-20 = 70°

           So

               I_1 = 700 \  cos^2 (70)

                   I_1= 81.884 W/m^2

 

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Let \sigma be the surface charge density of the of the non conducting parallel sheet.Let consider a Gaussian surface in the form of of cylinder such that its cross-sectional is A . Then there will be flux only due to cross sectional area as the curved sectional is perpendicular to the the electric field  so the Electric Flux due to it is zero.

Now using Gauss law we have, E be the electric Field at the distance r from the sheet then

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a) If we assume that the turntable is rotating at a constant speed, the only force acting on the seed parallel to the surface, which keeps it  from following a straight line trajectory, is the centripetal force.

So, we can apply Newton's 2nd Law to the seed in this way:

Fnet = m*a = m*ac = m*ω²*r

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As the seed does not slip, the only force acting  on it parallel to the surface, is the static  friction force, which has a maximum value, as follows:

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Ff = μ*m*g

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