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BartSMP [9]
3 years ago
13

Monochromatic light of wavelength 385 nm is incident on a narrow slit. On a screen 3.00 m away, the distance between the second

diffraction minimum and the central maximum is 1.40 cm. (a) Calculate the angle of diffraction θ of the second minimum.(b) Find the width of the slit.
Physics
1 answer:
LiRa [457]3 years ago
8 0

To solve this problem it is necessary to apply the concepts related to the concept of overlap and constructive interference.

For this purpose we have that the constructive interference in waves can be expressed under the function

a sin\theta = m\lambda

Where

a = Width of the slit

d = Distance of slit to screen

m = Number of order which represent the number of repetition of the spectrum

\theta = Angle between incident rays and scatter planes

At the same time the distance on the screen from the central point, would be

sin\theta = \frac{y}{d}

Where y = Represents the distance on the screen from the central point

PART A ) From the previous equation if we arrange to find the angle we have that

\theta = sin^{-1}(\frac{y}{d})

\theta = sin^{-1}(\frac{1.4*10^{-2}}{3})

\theta = 0.2673\°

PART B) Equation both equations we have

a sin\theta = m\lambda

a \frac{y}{d} = m\lambda

Re-arrange to find a,

a = \frac{(2)(385*10^{-9})(3)}{(1.4*10^{-2})}

a = 1.65*10^{-4}m

You might be interested in
When waves superimpose and make bigger amplitudes what form of interference is that
eimsori [14]

Answer:

Constructive Interference

Explanation:

Constructive Interference occurs when two waves superimpose and make bigger amplitudes.

In constructive interference, the crests of one wave fall on the crests of second wave and the amplitudes add up. The amplitude of the resultant wave is equal to sum of the amplitude of the individual waves. Similarly, the trough of first wave falls on the trough of other wave and they superimpose to create the trough of the resultant wave.

For Example, In the attachment, two waves A and B superimpose and demonstrate Constructive interference to create the wave C.

7 0
3 years ago
Measurement conversions [metric to metric] 568 cm = m​
Trava [24]

Answer:

5.68 meters

Explanation:

hope this helps!

7 0
3 years ago
what is the rotational kinetic energy of the earth? use the moment of inertia you calculated in part a rather than the actual mo
Ivenika [448]

The Earth's rotational kinetic energy is the kinetic Energy that the Earth

has due to rotation.

The rotational kinetic energy of the Earth is approximately <u>3.331 × 10³⁶ J</u>

Reasons:

<em>The parameters required for the question are;  </em>

<em>Mass of the Earth, M = </em><em>5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = </em><em>6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = </em><em>24.0 hrs</em><em>.</em>

The \ moment  \ of \  inertia \  of \  uniform \  sphere \  is \ I =   \mathbf{\dfrac{2}{5} \cdot M \cdot R^2}

Which gives;

\mathbf{I_{Earth}} =   \dfrac{2}{5} \times 5.97 \times 10 ^{24} \cdot \left(6.38 \times 10^6 \right)^2 = 9.7202107 \times 10^{37}

\mathrm{The \ rotational \  kinetic  \ energy \  is} \   E_{rotational} = \mathbf{\dfrac{1}{2} \cdot I \cdot \omega^2}

\mathrm{The \ angular \ speed, \ \omega} = \mathbf{\dfrac{2 \dcdot \pi}{T}}

Therefore;

\omega = \dfrac{2 \cdot \pi}{24}  = \dfrac{\pi}{24}

Which gives;

\mathbf{E_{rotational}} = \dfrac{1}{2} \times  9.7202107 \times 10^{37} \times  \left(  \dfrac{\pi}{12} \right)^2 = 3.331 \times 10^{36}

The rotational kinetic energy of the Earth, E_{rotational} = <u>3.331 × 10³⁶ Joules</u>

Learn more here:

brainly.com/question/13623190

<em>The moment of inertia from part A  of the question (obtained online) is that of the Earth approximated to a perfect sphere</em>.

<em>Mass of the Earth, M = 5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = 6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = 24.0 hrs</em>

3 0
3 years ago
If the half-life of iodine-131 is 8.10 days, how long will it take a 50.00 g sample to decay to 6.25 g?
Sloan [31]

Answer:

your in mr langfords class

Explanation:

bruh moment

6 0
3 years ago
The wavelength of the visible line in the hydrogen spectrum that corresponds to m = 5 in the Balmer equation is: A. 656 nm. B. 4
BaLLatris [955]

Answer:

The wavelength of the visible line in the hydrogen spectrum is 434 nm.

Explanation:

It is given that, the wavelength of the visible line in the hydrogen spectrum that corresponds to n₂ = 5 in the Balmer equation.

For Balmer series, the wave number is given by :

\dfrac{1}{\lambda}=R(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})

R is the Rydberg's constant

For Balmer series, n₁ = 2. So,

\dfrac{1}{\lambda}=1.097\times 10^7\times (\dfrac{1}{2^2}-\dfrac{1}{5^2})

\lambda=4.34\times 10^{-7}\ m

or

\lambda=434\ nm

So, the wavelength of the visible line in the hydrogen spectrum is 434 nm. Hence, this is the required solution.

6 0
3 years ago
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