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4 years ago

What is occurring when a light wave diffracts?

Physics

2 answers:

arsen [322]4 years ago

7 0

The answer is A just took the test

Elena L [17]4 years ago

3 0

Answer: The light wave is bending around an obstacle or through a barrier shows diffraction.

Explanation :

Light is one of the forms of energy. It enables us to see objects clearly. It shows many phenomena like the reflection of light, refraction, interference, diffraction etc.

The light rays bend when it passes through a narrow opening or obstacles. The essential condition for the diffraction is that the size of an obstacle should be comparable to the wavelength of light.

Hence, the correct option is (a) " The light wave is bending around an obstacle or through a barrier".

You might be interested in

53. A recently discovered planet has a mass four times as great as Earth's and a radius twice as large as Earth's. What will be

marusya05 [52]

Answer:

$g' = g = 9.81 m/s^2$

so gravity will be same as that of surface of earth

Explanation:

As we know that the acceleration due to gravity is given as

$g = \frac{GM}{R^2}$

here we have

$M = 4M_e$

$R = 2R_e$

we know that for earth we have

$g = 9.81 = \frac{GM_e}{R_e^2}$

now if the radius and mass is given as above

$g' = \frac{G(4M_e)}{(2R_e)^2}$

$g' = \frac{GM_e}{R_e^2}$

$g' = g = 9.81 m/s^2$

so gravity will be same as that of surface of earth

6 0

3 years ago

Two points are on a disk turning at constant angular velocity. One point is on the rim and the other halfway between the rim and

DerKrebs [107]

Answer:

The point on the rim

Explanation:

All the points on the disk travels at the same angular speed $\omega$ , since they cover the same angular displacement in the same time. Instead, the tangential speed of a point on the disk is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the point from the centre of the disk

As we can see, the tangential speed is directly proportional to the distance from the centre: so the point on the rim, having a larger r than the point halway between the rim and the axis, will have a larger tangential speed, and therefore will travel a greater distance in a given time.

3 0

4 years ago

An automobile traveling 50km/hr. It decelerates at 5 meter per second square. How long will it take to stop the car? How far wil

const2013 [10]

We know, a = v/t
Here, a = 5 m/s²
v = 50 km/h= 13.88 m/s

Substitute their values into the expression:
5 = 13.88 / t
t = 13.88/5
t = 2.78 sec

Now, we know, v = d/t
13.88 = d/2.78
d = 13.88 * 2.78
d = 38.53 meter

In short, Your Answers would be:
i) It will take 2.78 sec
ii) It will travel for 38.53 m after a brakes applied.

Hope this helps!

3 0

3 years ago

Calculate the energy of a photon having a wavelength in thefollowing ranges.(a) microwave, with λ = 50.00 cmeV(b) visible, with

IgorLugansk [536]

(a) $2.5\cdot 10^{-6}eV$

The energy of a photon is given by:

$E=\frac{hc}{\lambda}$

where

$h=6.63\cdot 10^{-34}Js$ is the Planck constant

$c=3\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength

For the microwave photon,

$\lambda=50.00 cm = 0.50 m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{0.50 m}=4.0\cdot 10^{-25} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-25}J}{1.6\cdot 10^{-19} J/eV}=2.5\cdot 10^{-6}eV$

(b) $2.5 eV$

For the energy of the photon, we can use the same formula:

$E=\frac{hc}{\lambda}$

For the visible light photon,

$\lambda=500 nm = 5 \cdot 10^{-7}m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-7} m}=4.0\cdot 10^{-19} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.5 eV$

(c) $2500 eV$

For the energy of the photon, we can use the same formula:

$E=\frac{hc}{\lambda}$

For the x-ray photon,

$\lambda=0.5 nm = 5 \cdot 10^{-10}m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-10} m}=4.0\cdot 10^{-16} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-16}J}{1.6\cdot 10^{-19} J/eV}=2500 eV$

6 0

3 years ago

A bird sitting on the limb of a tree is moving about 30 km/s with respect to the Sun. If the bird takes 1 second to drop down to

Alexandra [31]

Answer:

This faulty reasoning is best countered with Newton's First Law.

Explanation:

Newton's First Law states that every object or body will remain in its state, be it of rest or uniform motion, unless it is affected by a force.

This means that even while sitting, we are all moving at extreme speeds with respect to the Sun, of approximately 30 km/s (yes, including the bird and the worm), but we don't even notice because we are all moving at approximately the same speed, and we aren't being affected by a force strong enough that could change our state of being in said extreme speed with respect to the Sun (that is, unless our Earth suddenly stopped moving around the Sun!).

With that being said, when the bird drops down from the limb of the tree, it moves by "adding" a vertical speed from the tree to the ground, but, in reality, the bird is still moving at 30 km/s in a given axis with respect to the Sun, as well as the worm. Now, since the worm didn't "add" a speed to move away from the bird (or, at least, we assume that's what happened), the bird finally catches the worm.

To sum it all up: Yes, the worm still moves at 30 km/s with respect to the Sun, but the bird moves at the same speed, PLUS a given speed from the tree to the ground that helps it get to the worm.

7 0

3 years ago