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svet-max [94.6K]

3 years ago

7

Sunlight is reflected off of a puddle of water ahead of a driver. The index of refraction of the water is 1.333. If a driver sit

s 1.5 m above the ground, at what distance from the driver will polarized sunglasses be most effective at blocking the reflected light from the puddle. Hint: What angle does light become maximally polarized? 5. A. 1.1m B. 3.0m C. 2.0 m D. 0.75 m E. 2.5m

1 answer:

Lynna [10]3 years ago

5 0

Answer:

option C is correct

distance from the driver is 2 m

Explanation:

Given data

index of refraction of the water n = 1.333

driver height from ground= 1.5 m

to find out

What angle does light become maximally polarized and what distance from the driver will polarized

solution

we know that n1 for water = 1.33 and we know for air n2 = 1

so tan(θ) = n1 / n2

tan(θ) = 1.33 / 1

so (θ) = 53.06 degree

so for distance d will be calculate as

tan ( 90 - (θ) ) = height / distance

tan ( 90 - 53.06 ) = 1.5 / distance

distance = 1.5 / 0.7519

distance = 1.994 m = 2 m

so option C is correct

distance from the driver is 2 m

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3 years ago

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If stellar parallax can be measured to a precision of about 0.01 arcsec using telescopes on the Earth to observe stars, to what

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3 years ago

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Complete Question:

Given $\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$ at a point. What is the force per unit area at this point acting normal to the surface with $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$ ? Are there any shear stresses acting on this surface?

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There are shear stresses acting on the surface since $\tau \neq 0$

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$\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$

equation of the normal, $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$

$\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

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$T_n = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

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$T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z$

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$\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa$

If the shear stress, $\tau$ , is calculated and it is not equal to zero, this means there are shear stresses.

$\tau = T_n - \sigma_n \b n$

$\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau = \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z$

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Since $\tau \neq 0$ , there are shear stresses acting on the surface.

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