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MaRussiya [10]
2 years ago
9

An instructor wishes to determine the wavelength of the light in a laser beam. To do so, he directs the beam toward a partition

with two tiny slits separated by 0.195 mm. An interference pattern appears on a screen that lies 5.10 m from the slit pair. The instructor's measurements show that two adjacent bright interference fringes lie 1.55 cm apart on the screen. What is the laser's wavelength (in nm)
Physics
1 answer:
Morgarella [4.7K]2 years ago
5 0

Answer:

λ = 610.6 nm

Explanation:

We are given;

Separation distance; d = 0.195 mm = 0.195 × 10^(-3) m

Interference pattern distance; D = 4.95 m

Width of the two adjacent bright interference; β = 1.55 cm = 1.55 × 10^(-2) m

Formula for the Fringe width is;

β = Dλ/d

Where;

λ is laser's wavelength

Thus;

λ = (d × β)/(D)

λ = (0.195 × 10^(-3) × 1.55 × 10^(-2))/4.95

λ = 610.6 × 10^(-9) m

λ = 610.6 nm

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Iteru [2.4K]

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6 0
2 years ago
Why is it important to know the location of active and inactive faults in our country?
Andrew [12]

Answer:

I hope this helps.

Explanation:

It's important to know the location of an active fault in order to determine the magnitude of the expected earthquake. There is a chance than an inactive fault can become active again. It's important that we take the locations into account in order to be prepared and ready for if it occurs.

5 0
3 years ago
On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = go at the north pole
ohaa [14]
The reason why there is a difference between free-fall acceleration is a centrifugal force.
I attached a diagram that shows how this force aligns with the force of gravity.
From the diagram we can see that:
F=F_g-F_{cf}=mg'-mw^2r'cos(\alpha)\\ ma=mg'-mw^2r'cos(\alpha)\\ a=g'-w^2rcos^2(\alpha)\\
Where g' is the free-fall acceleration when there is no centrifugal force, r is the radius of the planet, and w is angular frequency of planet's rotation. \alpha is the latitude.
We can calculate g' and wr^2 from the given conditions in the problem.
g(90)=g_0;\ g_0= g'-w^2rcos^2(90)\\
g_0=g'\\
g(0)=ag_0;\ ag_0=g_0-w^2rcos^2(0)\\
ag_0=g_0-w^2r\\
w^2r=g_0(a-1)

Our final equation is:
g=g_0-g_0(a-1)cos^2(\alpha)
Colatitude is:
\alpha_c=90^\circ-\alpha
The answer is:
g=g_0-g_0(a-1)cos^2(90-9)=g_0-g_0(a-1)sin^2(9)

5 0
3 years ago
A student pushed a box 32.0 m across a smooth, horizontal floor using a constant force of 124 N. If the force was applied for 8.
Brilliant_brown [7]

The power developed is 500 W ( to the nearest Watt)

Power(P) is the rate at which work is done. Work done (W) is the product of the force applied on the object and the displacement (s) made by the point of application of the force.

P = \frac{W}{t}

W= F*s

Therefore,

P=\frac{F*s}{t}

Substitute the given values of force , displacement and time

F = 124 N,s = 32.0 m,t = 8.0 s

P =\frac{W*s}{t} =\frac{124N*22.0s}{8.0s} =496 W

Thus the Power can be rounded off to the nearest value of 500 W

3 0
3 years ago
Read 2 more answers
An object is projected with initial speed v0 from the edge of the roof of a building that has height H. The initial velocity of
Vsevolod [243]

Answer:

Explanation:

Initial velocity u = V₀ in upward direction so it will be negative

u = - V₀

Displacement s = H . It is downwards so it will be positive

Acceleration = g ( positive as it is also downwards )

Using the formula

v² = u² + 2 g s

v² = (- V₀ )² + 2 g H

= V₀² + 2 g H .

v = √ ( V₀² + 2 g H )

6 0
3 years ago
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