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svet-max [94.6K]
2 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
Physics
1 answer:
Lynna [10]2 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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Answer:

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Explanation:

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3 years ago
A camera with a 50.0-mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens m
kirill [66]

a) 50.8 mm

b) The whole image (1:1)

c) It seems reasonable

Explanation:

a)

To project the image on the film, the distance of the film from the lens must be equal to the distance of the image from the lens. This can be found by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length of the lens

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem:

f = 50.0 mm = 0.050 m is the focal length (positive for a convex lens)

p = 3.00 m is the distance of the person from the lens

Therefore, we can find q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{0.050}-\frac{1}{3.00}=19.667m^{-1}\\q=\frac{1}{19.667}=0.051 m=50.8 mm

b)

Here we need to find the height of the image first.

This can be done by using the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where:

y' is the height of the image

y = 1.75 m is the height of the real person

q = 50.8 mm = 0.0508 m is the distance of the image from the lens

p = 3.00 m is the distance of the person from the lens

Solving for y', we find:

y'=-\frac{qy}{p}=-\frac{(0.0508)(1.75)}{3.00}=-0.0296 m=-29.6mm

(the negative sign means the image is inverted)

Therefore, the size of the image (29.6 mm) is smaller than the size of the film (36.0 mm), so the whole image can fit into the film.

c)

This seems reasonable: in fact, with a 50.0 mm focal length, if we try to take the picture of a person at a distance of 3.00 m, we are able to capture the whole image of the person in the photo.

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2 years ago
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luda_lava [24]
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6 0
3 years ago
Read 2 more answers
The maximum distance which our normal eye can see distinctly is known as _______________
azamat

Answer:

Far point.

Explanation:

The maximum distance up to which the normal eye can see objects distinct and clear is called the far point of the eye. It is infinity for a normal eye.

7 0
2 years ago
Jeff of the Jungle swings on a 7.6-m vine that initially makes an angle of 32 ∘ with the vertical.
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To solve this problem we will use the trigonometric concepts to find the distance h, which will allow us to find the speed of Jeff and that will finally be the variable that will indicate the total tension, since it is the variable of the centrifugal Force given in the vine at the lowest poing of the swing.

From the image:

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v = \sqrt{2gh}

v = \sqrt{2(9.8)(1.1548)}

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T = W+F_c

T = mg + \frac{mv^2}{r}

T = (83)(9.8)+\frac{(9.8)( 4.75)^2}{7.6}

T = 842.49N

Therefore the tension in the vine at the lowest point of the swing is 842.49N

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