Can someone help me with this please?

When a certain element is excited with electricity, we see three main lines in its emission spectrum: two red lines and one oran

Eddi Din [679]

The absorption spectrum would have all the wavelengths of the light source but would have black lines where the two red and one orange lines were in the emission spectrum

4 0

3 years ago

What does this indicate about the claim?

Anton [14]

claim

[klām]

VERB

state or assert that something is the case, typically without providing evidence or proof.

"he claimed that he came from a wealthy, educated family" · [more]

synonyms:

assert · declare · profess · maintain · state · hold · affirm · avow · aver · [more]

NOUN

an assertion of the truth of something, typically one that is disputed or in doubt.

"he was dogged by the claim that he had CIA links" · [more]

synonyms:

assertion · declaration · profession · affirmation · avowal · averment · protestation · representation · contention · submission · case · allegation · pretense · asseveration

a demand or request for something considered one's due.

"the court had denied their claims to asylum"

7 0

3 years ago

Read 2 more answers

In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo

Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of $32^{\circ}$ (when light enters from the air.)

Explanation:

Let $v_{1}$ denote the speed of light in the first medium. Let $v_{\text{air}}$ denote the speed of light in the air. Assume that the light entered the boundary at an angle of $\theta_{1}$ to the normal and exited with an angle of $\theta_{\text{air}}$ . By Snell's Law, the sine of $\theta_{1}\!$ and $\theta_{\text{air}}\!$ would be proportional to the speed of light in the corresponding medium. In other words:

$\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}$ .

When light enters a boundary at the critical angle $\theta_{c}$ , total internal reflection would happen. It would appear as if the angle of refraction is now $90^{\circ}$ . (in this case, $\theta_{\text{air}} = 90^{\circ}$ .)

Substitute this value into the Snell's Law equation:

$\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}$ .

Rearrange to obtain an expression for the speed of light in the first medium:

$v_{1} = v_{\text{air}} \cdot \sin(\theta_{1})$ .

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For $0 < \theta < 90^{\circ}$ , $\sin(\theta)$ is monotonically increasing with respect to $\theta$ . In other words, for $\!\theta$ in that range, the value of $\sin(\theta)\!$ increases as the value of $\theta\!$ increases.

Therefore, compared to the medium in this question with $\theta_{c} = 27^{\circ}$ , the medium with the larger critical angle $\theta_{c} = 32^{\circ}$ would have a larger $\sin(\theta_{c})$ . such that light would travel faster in that medium.