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Sergeu [11.5K]
3 years ago
8

What is the wavelength of light that is deviated in the first order through an angle of 13.1 ∘ by a transmission grating having

5000 slits/cm? Assume normal incidence.
Physics
1 answer:
faltersainse [42]3 years ago
7 0

Answer:

The wavelength of light is 4.53\times10^{-7}\ m

Explanation:

Given that,

Angle = 13.1°

Number of slits = 5000

We need to calculate the wavelength of light

Diffraction of first order is defined as,

d \sin\theta=n\lambda.....(I)

The separation of the slits

d = \dfrac{1}{N}

d=\dfrac{1}{5000}

d=2\times10^{-6}\ m

Now put the value in equation (I)

2\times10^{-6}\sin13.1^{\circ}=\lambda

Here, n = 1

\lambda=4.53\times10^{-7}\ m

Hence, The wavelength of light is 4.53\times10^{-7}\ m

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A hole of radius r is bored through the center of a sphere of radius r. Find the volume v of the remaining portion of the sphere
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Answer:

Πr²(4r/3 - h)

Explanation:

Volume of a sphere is 4/3Πr³. If a hole of radius r is bored through, the hole with generate a circular shape in the sphere. The volume of the remaining portion of the sphere will be the difference between the volume of the sphere and the area of the hole bored(which will be volume of a cylinder since the hole bored will create a cylindrical shape in the sphere)

Area of the remaining portion = Volume of sphere - volume of a cylinder

Volume of sphere = 4/3Πr³

Volume of a cylinder = Πr²h

Volume of the remaining portion = 4/3Πr³ - Πr²h

= Πr²(4r/3 - h)

Where h is the height of the cylindrical hole

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2 years ago
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Why does time slow down the closer you get to a black hole, I know that time slows down the faster you move through space and th
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Physicists and engineers from around the world have come together to build the largest accelerator in the world, the Large Hadro
elena-14-01-66 [18.8K]

Solution :

Energy of photon, E = 6.7 eV

                              E = $6.7 \times 1.602 \times 10^{-7}$ joule

Kinetic energy, $K.E. =\frac{1}{2} mv^2 = 1.602 \times 6.7 \times 10^{-7}$

$v^2=\frac{2 \times 1.602 \times 6.7 \times 10^{-7}}{1.6726 \times 10^{-27}}$

   $=12.834 \times 10^{-20}$

Kinetic energy at high speeds

$(r-1)\times mc^2 = 6.7 \ eV$

$(r-1)=\frac{6.7 \times 1.602 \times 10^{-7}}{1.6726 \times 10^{-27} \times 9 \times 10^{16}}$

r - 1 = 7130

r = 7130 + 1

r  = 7131

$\frac{1}{\sqrt{1-\frac{v^2}{C^2}}}=7131$

$1-\frac{v^2}{C^2} = \left(\frac{1}{7131}\right)^2$

$v^2=C^2\left[1-\left(\frac{1}{7131}\right)^2\right]$

$v=0.99999999017C$

Δ = 1 - 0.99999999017

   = 0.00000000933

Relative mass, $m_{rel}=r.m$

                                $=7131 \times 1.6728 \times 10^{-27}$

                               $=1.1927 \times 10^{-23}$ kg

                                 

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He ordered into increasing atomic mass
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