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2 years ago

Please help meh need answers

Physics

1 answer:

boyakko [2]2 years ago

4 0

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:What will be the value of the refractive index of the medium? Critical angle of that medium is 30 degree

earnstyle [38]

Answer:

Let the second medium be air (n₁=1)

The refractive index n₂ of the medium where first medium is air is found (a)

(a) n₂ = 2

Explanation:

Critical angle can be defined as the angle of incidence that provides the angle of refraction of 90°.

Refractive index of a medium can be defined as a number that describes that how fast a light will travel through that medium.

Critical angle and Refractive index are related by:

$\theta_{critical}= sin^{-1}(\frac{n_1}{n_2})$

$sin \theta_{critical}=\frac{n_1}{n_2}$

To find refractive index of medium with respect to air, substitute n₁=1 (Refractive index of air is 1)

Also θ(critical)=30°

Find n₂ :

$sin30= \frac{1}{n_2}\\0.5=\frac{1}{n_2}\\n_2=\frac{1}{0.5}\\n_2=2$

8 0

2 years ago

Consider the hydrogen atom. How does the energy difference between adjacent orbit radii change as the principal quantum number i

Kisachek [45]

Answer:

the energy difference between adjacent levels decreases as the quantum number increases

Explanation:

The energy levels of the hydrogen atom are given by the following formula:

$E=-E_0 \frac{1}{n^2}$

where

$E_0 = 13.6 eV$ is a constant

n is the level number

We can write therefore the energy difference between adjacent levels as

$\Delta E=-13.6 eV (\frac{1}{n^2}-\frac{1}{(n+1)^2})$

We see that this difference decreases as the level number (n) increases. For example, the difference between the levels n=1 and n=2 is

$\Delta E=-13.6 eV(\frac{1}{1^2}-\frac{1}{2^2})=-13.6 eV(1-\frac{1}{4})=-13.6 eV(\frac{3}{4})=-10.2 eV$

While the difference between the levels n=2 and n=3 is

$\Delta E=-13.6 eV(\frac{1}{2^2}-\frac{1}{3^2})=-13.6 eV(\frac{1}{4}-\frac{1}{9})=-13.6 eV(\frac{5}{36})=-1.9 eV$

And so on.

So, the energy difference between adjacent levels decreases as the quantum number increases.

5 0

2 years ago

According to the concept of length contraction, what happens to the length of an object as it approaches the speed of light and

lord [1]

When an object moves its length contracts in the direction of motion. The faster it moves the shorter it gets in the direction of motion.
The object in this question moves and then stops moving. So it's length first contracts and then expands to its original length when the motion stops.
The speed doesn't have to be anywhere near the speed of light. When the object moves its length contracts no matter how fast or slow it's moving.

8 0

3 years ago

Free runners jump long distances and land on the ground or a wall. How do they apply Newton’s second law to lessen the force of

Veseljchak [2.6K]

As we know that as per Newton's II law we have

$F = \frac{dP}{dt}$