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Light of wavelength 559 nm is used to illuminate normally two glass plates 22.1 cm in length that touch at one end and are separ

umka21 [38]

Answer:

M = 222 fringes

Explanation:

given

λ = 559 n m = 559 × 10⁻⁹ m

radius = 0.026 mm = 0.026 ×10⁻³ m

length of the glass plate = 22.1 ×10⁻² m

using relation

$2t=(m+\dfrac{1}{2})\lambda\ \ (m=0,1,2,3...)\\where\ 0\leq t\leq 2r\\m = \dfrac{2t}{\lambda}-\dfrac{1}{2}$

$m_{max} = \dfrac{2\times 2r}{\lambda}-\dfrac{1}{2}\\m_{max} = \dfrac{2\times 2\times 0.026\times 10^{-3}}{559\times 10^{-9}}-\dfrac{1}{2}$

= 221.79

= 221 (approx.)

hence no of bright fringe

M = m + 1

= 221 +1

M = 222 fringes

6 0

3 years ago

What is the Deepest point of the ocean and where is it?

qaws [65]

Answer:

marinas trench in Pacific Ocean

<h2>mark it as brainliest</h2>

5 0

2 years ago

Read 2 more answers

Transition

Snezhnost [94]

Frozen water has move volume than water in liquid form

8 0

3 years ago

At time t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is (7.00 m/s2)i hat + (6.00 m/s2)j. It m

Elina [12.6K]

Answer:

r = 3519.55 m

Explanation:

We know that the acceleration of a particle in a circular motion is directed towards the center of the circumference and has magnitude:

F = rω^2

Where r is the radius of the circumference and ω is the angular velocity.

From the two acceleration vectors we find that their magnitude is

√(7^2+6^2) = √85

Therefore:

√85 m/s^2= rω^2

Now we need to calculate the angular velocity to obtain the radius. Since t2-t1 = 3s is less than one period we can be sure that the angular velocity is equals to the angle traveled between this time divided by 3 s.

The angle with respect to the x-axis for the particle at t1 and t2 is:

$\theta 1 =\cos ^{-1}\left(\frac{7}{\sqrt{85}}\right)\\\theta 2 =\cos ^{-1}\left(\frac{6}{\sqrt{85}}\right)\\$

Therefore, the angular velocity ω is (in radians per second):

$\omega = \frac{\theta2 - \theta1}{3 s} = 0.0511813 \frac{1}{s}$

Therefore:

r = √85 / (0.0511813)^2 = 3519.55 m

4 0

3 years ago

Two lasers are shining on a double slit, with slit separationd.Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelengt

Sophie [7]

Answer:

a) laser 1 has the maximum closest to the central maximum

b) y₂ –y₁ = L 1.66 10⁻²

Explanation:

a), B1, B2) The expression that describes the constructive interference for a double slit is

d sin θ = m λ

The pattern is observed on a screen

tan θ = y / L

Since the angles are very small

tan θ = sin θ / cos θ = sin θ = y/L

d y / L = m λ

In this case the laser has a wavelength

λ ₁ = d/20

We substitute

d y / L = m d / 20

m = 1

y₁ = L / 20

For the laser 2 λ ₂= d / 15

y₂ = L / 15

When examining the two expressions, laser 1 has the maximum closest to the central maximum

b) the difference between the two patterns is

y₂- y₁ = L (1/15 - 1/20)

y₂ –y₁ = L 1.66 10⁻²

C) laser 1 second maximum

y₁ ’= 2 L / 20

y₁ ’= L 0.1

Laser 2 third minimum

To have a minimum, the equation must be satisfied

d sin θ = (m + ½) λ

d y / L = (m + ½) λ

d y / L = (m + ½) d / 15

y = L (m +1/2) / 15

m = 3

y₂’= L (3 + ½) / 15

y₂’= L 0.2333

The difference is

y₁ ’- y₂’ = L (0.1 - 0.2333)

y₁ ’–y₂’ = L (-0.133)

3 0

3 years ago

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