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2 years ago

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The index of refraction of a certain material is 1.25. If I send red light (700 nm) through the material, what will the frequenc

y of the light be in the material

1 answer:

Readme [11.4K]2 years ago

6 0

Answer:

f1 / f2 = n2 / n1

Explanation:

To solve this problem, we should remember that the formula for index of refraction is defined as:

n = c / v

or

n v = c

Where,

n = index of refraction

c = speed of light

v = speed of light in the medium

Since speed of light is constant, then we can simply equate the materials 1 and 2:

n1 v1 = n2 v2

Where the speed of light in the medium (v) can be expressed as:

v = w * f

Where,

w = wavelength of light

f = frequency of light

Therefore substituting this back into the relating equation:

n1 w1 f1 = n1 w2 f1

Since it is given that the light is monochromatic, w1 = w2, this further simplifies the equation to:

n1 f1 = n2 f2

f1 / f2 = n2 / n1 (ANSWER)

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Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

2 years ago

At what distance above earth would a satellite have a period of 125 min?

Nezavi [6.7K]

Rw^2 = GmM/r^2
<span> Leads to
</span><span> w^2 r^3 = GM
</span><span> (2pi /T) ^2 r^3 = GM
</span><span> 4pi^2 r^3 = GM T^2
</span><span> r^3 = GM T^2 / 4pi^2
</span><span> Work out r^3 then r.
</span> T = 125 min = 125(60) = 7500 s
<span> R = 6.38E6 m
</span><span> m = 5.97E24 kg
</span><span> G = 6.673E-11
</span> r=<span> 8279791.78</span><span> m
Since r = radius R of Earth + height above urface,h
</span><span> h = r - R = </span><span> 8279791.78 - </span>6.38E6 = <span> <span>1899791.78 m
h=</span></span><span> <span>1899.79178 Km</span></span>

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2 years ago

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Who pays for Government workers that work on alcohol impaired driving cases?

Shalnov [3]

Answer: Taxpayers

Explanation:

Taking alcohol before driving or while driving is dangerous and has resulted in lots of accidents and deaths. Alcohol tampers with the normal functioning of the brain, and also impairs ones reasoning.

Alcohol impaired driving cases handled by government officials are paid for by the taxpayers. A tax is the levy that the people in the country pays. Those funds are used in handling different government objectives and this is one of such ways.

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1 year ago

a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres

shusha [124]

Answer:

Part a) When collision is perfectly inelastic

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b) When collision is perfectly elastic

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

$mv_m = (m + M)v$

so we have

$v = \frac{mv_m}{m + M}$

now we know that in order to complete the circle we will have

$v = \sqrt{5Rg}$

$\frac{mv_m}{m + M} = \sqrt{5Rg}$

now we have

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b)

Now we know that collision is perfectly elastic

so we will have

$v = \frac{2mv_m}{m + M}$

now we have

$\sqrt{5Rg} = \frac{2mv_m}{m + M}$

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

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2 years ago

2. Heavier football players tend to play on the front line. Why? <br> What law is it?

gogolik [260]

Answer: They are put in front for defense so so they can block the opponents from getting the ball

Explanation:

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1 year ago

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