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olga nikolaevna [1]
2 years ago
13

Light with a wavelength of about 490 nm is made to pass through a diffraction grating. the angle formed between the path of the

incident light and the diffracted light is 9.2° and forms a first-order bright band. what is the number of lines per mm in the diffraction grating? round your answer to the nearest whole number. lines per mm
Physics
1 answer:
polet [3.4K]2 years ago
5 0

The number of lines per mm in the diffraction grating is 326.

<h3>What is diffraction grating?</h3>

A diffraction grating is a type of optical instrument obtained with a continuous pattern. The pattern of the diffracted light by a grating depends on the structure and number of elements present.

The given data in the problem is

\rm \theta is the angle formed between the path of the incident light and the diffracted light = 9. 2°

λ is the wavelength of the light=490nm=4.9

N is the number of lines per mm in the diffraction grating=?

n is ordered = 1

The formula for the diffraction grating is;

n \lambda = d sin\theta \\\\ d = \frac{n \lambda}{sin \theta } \\\\  d = \frac{1 \times 4.90 \times 10^{-7}}{sin 9.2^0 } \\\\ d=3.06 \times 10^{-6} \\\\ d=3.06 \times 10^{-3} \ mm

The number of lines per mm is found as;

\rm N= \frac{1}{d} \\\\ N= \frac{1}{3.06 \TIMES 10^{-3}} \\\\ N=326.8 /mm

Hence the number of lines per mm in the diffraction grating is 326.

To learn more about diffraction grating refer to the link;

brainly.com/question/1812927

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AysviL [449]
K.E. = 1/2 mv²
K.E. is directly proportional to v^2
So, when K.E. increase by 2, K.E. increase by root. 2
v' = 1.41v
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After round-off to it's tenth value, it will be:
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So, option B is your answer!

Hope this helps!

7 0
3 years ago
A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a
Elza [17]

Answer:

\dfrac{dz}{dt}=0.65\ ft/s

Explanation:

Given that

x= 150 ft

\dfrac{dy}{dt}= 7\ ft/s

y= 14 ft

From the diagram

z^2=x^2+y^2

When ,x= 150 ft and y= 14 ft

z^2=150^2+14^2

z=\sqrt{150^2+15^2}

z=150.74 ft

z^2=x^2+y^2

By differentiating with respect to time t

2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}

z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}

Here x is constant that is why

\dfrac{dx}{dt}=0

z\dfrac{dz}{dt}= y\dfrac{dy}{dt}

Now by putting the values in the above equation we get

150.74\times \dfrac{dz}{dt}=14\times 7

\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s

\dfrac{dz}{dt}=0.65\ ft/s

Therefore the distance between balloon and observer increasing with 0.65 ft/s.

5 0
3 years ago
By what factor must the amplitude of a sound wave be decreased in order to decrease the intensity by a factor of 3?
s344n2d4d5 [400]

Answer:

Amplitude is decreased by a factor of \sqrt3 if intensity is decreased by a factor of 3.

Explanation:

Intensity of a sound wave is directly proportional to the square of its amplitude.

Therefore, if intensity is I and amplitude is A, then

I=kA^2, where, k is constant of proportionality.

Now, if intensity of sound wave is decreased by a factor of 3. So,

New intensity is, I_{new}=\frac{I}{3}

I_{new}=kA_{new}^2\\\frac{I}{3}=kA_{new}^2

Plug in kA^2 for I. This gives,

\frac{kA^2}{3}=kA_{new}^2\\A_{new}^2=\frac{A^2}{3}\\A_{new}=\sqrt{\frac{A^2}{3}}=\frac{A}{\sqrt{3}}

Therefore, amplitude is decreased by a factor of \sqrt3.

4 0
3 years ago
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erastova [34]

Answer:

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total time in air = 2 * 3.47 sec = 6.94 sec

(time rising must equal time falling)

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Can also use range formula

R = v^2 sin (2 theta) / g

tan theta = 34 / 17 = 2

theta = 63.4 deg

2 theta = 126.9 deg

sin 126.9 = .8

v^2 = 17^2 + 34^2 = 1445 m^2/s^2

R = 1445 * .8 / 9.8 = 118 m    agreeing with answer found above

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(credit to google) 

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