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ratelena [41]
3 years ago
15

The width of the central maxima, formed from light of wavelength 575 nm behind a single slit that has a width of 115 μm, is 1.15

cm, what is the distance between the slit and the screen?
Physics
1 answer:
Lady bird [3.3K]3 years ago
3 0

Answer:

 L  = 1.15 m

Explanation:

The diffraction phenomenon is described by the equation

        a sin θ = m λ

Where a is the width of the slit, λ  the wavelength and m is an integer, the order of diffraction is left.

The diffraction measurements are made on a screen that is far from the slit, and the angles in the experiment are very small, let's use trigonometry

          tan θ = y / L

          tan θ = sint θ / cos θ≈ sin θ

We substitute in the first equation

           a (y / L) = m λ

The first maximum occurs for m = 1

The distance is measured from the center point of maximum, which coincides with the center of the slit, in this case the distance is the total width of the central maximum, so the distance (y) measured from the center is

         y = 1.15 / 2 = 0.575 cm

         y = 0.575 10⁻² m

Let's clear the distance to the screen (L)

       L = a y / λ  

Let's calculate

     L = 115 10⁻⁶  0.575 10⁻² / 575 10⁻⁹

     L  = 1.15 m

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A car travelling at 15 m/s comes to rest in a distance of 14 m when the brakes are applied.
stira [4]

Answer:

-8.04 m/s2

Explanation:

To find the answer to this, you have to use the 4th kinematic equation:

v^{2} = v^{2}_{0}  + 2ax

You plug into the equation to get:

0 = 15^{2} + 2a(14)

solve for a to get

-8.04 m/s2

3 0
3 years ago
Which of the following are examples of projectile motion?
liubo4ka [24]

Answer: A. a basketball being shot toward the basket

Explanation: The definition of projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. So, the basketball is the object being thrown and the person throwing the ball is aiming it to go into the basket making that the path of trajectory. Hope that makes sense and helps!

5 0
2 years ago
4
Zigmanuir [339]

(a) The object moves with uniform velocity from A to B.

(b) The object moves with constant velocity from B to C.

(c) The object moves with increasing velocity from C to D.

<h3>Velocity of the object from point A to B</h3>

V(A to B) = (6 - 0)/(4 - 0) = 1.5 m/s

<h3>Velocity of the object from point B to C</h3>

V(B to C) = (6 - 6)/(11 - 4) = 0 m/s

<h3>Velocity of the object from point C to D</h3>

V(C to D) = (7 - 6)/(12 - 11) = 1 m/s

final velocity = 1 + 1.5 m/s = 2.5 m/s

Thus, we can conclude the following;

The object moves with uniform velocity from A to B.

The object moves with constant velocity from B to C.

The object moves with increasing velocity from C to D.

Learn more about velocity here: brainly.com/question/6504879

#SPJ1

4 0
2 years ago
Consider a horizontal layer of the dam wall of thickness dx located a distance x above the reservoir floor. What is the magnitud
adoni [48]

Answer:

Explanation:

Attached is the solution

6 0
3 years ago
A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal
Gemiola [76]

Answer:

Time taken, T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

Explanation:

It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.

From the figure,

The sum of forces in y direction is :

T\ cos\theta-mg=0

T=\dfrac{mg}{cos\theta}

Sum of forces in x direction,

T\ sin\theta=\dfrac{mv^2}{r}

mg\ tan\theta=\dfrac{mv^2}{r}.............(1)

Also, r=l\ sin\theta

Equation (1) becomes :

mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}

v=\sqrt{gl\ tan\theta.sin\theta}...............(2)

Let t is the time taken for the ball to rotate once around the axis. It is given by :

T=\dfrac{2\pi r}{v}

Put the value of T from equation (2) to the above expression:

T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}

T=\dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta.sin\theta}}

On solving above equation :

T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

Hence, this is the required solution.

4 0
3 years ago
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