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1 year ago

450 nm of light falls on a single slit of width 0.30 mm. What is the angular width of the central diffraction peak

Physics

1 answer:

Andre45 [30]1 year ago

4 0

Angular width is 3 x 10^-3

Let D be the distance between source and screed d the distance between coherent source then for central diffraction maxima,

where λ is wavelength

Given:

λ = 450 nm = 450×10^−9m

d = 0.3x10^−3m, D = 1m

W = 2 x 450×10^−9/0.3x10^−3*1

To Find:

Angular width

Solution: The width of the central maxima is nothing but the difference between the positions of the first two minima. Hence we will use the expression for the position of minima and accordingly obtain the expression of the width of central maxima and secondary maxima

θ = W/D

θ = 2 x 450×10^−9/0.3x10^−3*1/1 = 3 x 10^-3

Hence, angular width is 3 x 10^-3

Learn more about Angular width here:

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stich3 [128]

Answer:

Electric field acting on the electron is 127500 N/C.

Explanation:

It is given that,

Mass of an electron, $m=9.11\times 10^{-31}\ kg$

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Initial speed of electron, u = 0

Final speed of electron, $v=3\times 10^7\ m/s$

Distance covered, s = 2 cm = 0.02 m

We need to find the electric field required. Firstly, we will find the acceleration of the electron from third equation of motion as :

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(3\times 10^7)^2-0}{2\times 0.02}$

$a=2.25\times 10^{16}\ m/s^2$

According to Newton's law, force acting on the electron is given by :

F = ma

$F=9.1\times 10^{-31}\times 2.25\times 10^{16}$

$F=2.04\times 10^{-14}\ N$

Electric force is given by :

F = q E, E = electric field

$E=\dfrac{F}{q}$

$E=\dfrac{2.04\times 10^{-14}}{1.6\times 10^{-19}}$

E = 127500 N/C

So, the electric field is 127500 N/C. Hence, this is the required solution.

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3 years ago

A force of 150 N is applied to a 25 cm- piston in a hydraulic machine. What is the area

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Answer:

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2 years ago

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<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Similar charges repel each other, whereas charges that are opposed attract each other.

Given data;

Electric force,F = 9 × 10 ⁻⁹ N

Distance between charges,d = 7 × 10⁻⁴ m

Chrge,q₁ = q₂ =q C

From Columb's law;

$\rm F = K \frac{q_1q_2}{d^2} \\\\ 9 \times 10^{-9} = 9 \times 10^9 \frac{q^2}{(7 \times 10^{-4})^2} \\\\ q^2 = 4.9 \times 10^{-25} \\\\ q = 7 \times 10^{-13} \ C$