For the scenarios in which total internal reflection is possible, rank the scenarios on the basis of the critical angle, the angle above which total internal reflection occurs. At this angle, the refracted ray is at 90 degrees from the normal. Rank from largest to smallest. To rank items as equivalent, overlap them.

ITEM A: n1benzene =1.50 n2water =1.33

ITEM B: n1diamond =2.42 n2water =1.33

ITEM C: n1diamond =2.42 n2air =1.00

ITEM D: n1water =1.33 n2air =1.00

Let [tex]n_i[/tex] be the index of refraction for the medium where the light comes from.
Let [tex]n_r[/tex] be the index of refraction of the medium which the light enters after refraction.

Note that to achieve total internal reflection, [tex]\displaystyle \frac{n_r}{n_i} \le 1[/tex]. In other words, [tex]n_r \le n_i[/tex]. That's a hint for distinguishing between the two media.

By Snell's Law,

[tex]n_i\, \sin \theta_i = n_r\, \sin \theta_r[/tex].

Rearrange this equation to obtain:

[tex]\displaystyle \sin \theta_i = \frac{n_r\, \sin \theta_r}{n_i}[/tex].

At the smallest [tex]\theta_i[/tex] that ensures total internal reflection, [tex]\theta_r = 90^\circ[/tex].

Accordingly, [tex]\sin \theta_r = 1[/tex]. Hence, the equation from Snell's Law would become

[tex]\displaystyle \sin \theta_i = \frac{n_r\, \sin \theta_r}{n_i} = \frac{n_r}{n_i}[/tex].

The angle of incidence [tex]\theta_i[/tex] should be between [tex]0^\circ[/tex] and [tex]90^\circ[/tex]. On that range, [tex]\sin \theta_i[/tex] would be strictly increasing. To compare the

In this question,

Item A: [tex]\displaystyle \frac{n_r}{n_i} = \frac{1.33}{1.50} \approx 0.887[/tex].
Item B: [tex]\displaystyle \frac{n_r}{n_i} = \frac{1.33}{2.42} \approx 0.550[/tex].
Item C: [tex]\displaystyle \frac{n_r}{n_i} = \frac{1.00}{2.42} \approx 0.413[/tex].
Item D: [tex]\displaystyle \frac{n_r}{n_i} = \frac{1.00}{1.33} \approx 0.752[/tex].

Rank of [tex]\displaystyle \frac{n_r}{n_i}[/tex] from largest to smallest: [tex]\rm A > D > B > C[/tex].

[tex]\implies[/tex]Rank of [tex]\sin \theta_i[/tex] from largest to smallest [tex]\rm A > D > B > C[/tex].

[tex]\implies[/tex]Rank of [tex]\theta_i[/tex] from largest to smallest [tex]\rm A > D > B > C[/tex].

Madisonchurchwell5 Madisonchurchwell5 · Answer 2 · 2025-03-27T15:24:21-04:00

Answer:

(no overlap.)

Explanation:

Let

be the angle of incidence, and

be the angle of refraction.

Also:

Let be the index of refraction for the medium where the light comes from.

Let be the index of refraction of the medium which the light enters after refraction.

Note that to achieve total internal reflection, . In other words, . That's a hint for distinguishing between the two media.

By Snell's Law,

.

Rearrange this equation to obtain:

.

At the smallest that ensures total internal reflection, .

Accordingly, . Hence, the equation from Snell's Law would become

.

The angle of incidence should be between and . On that range, would be strictly increasing. To compare the

In this question,

Item A: .

Item B: .

Item C: .

Item D: .

Rank of from largest to smallest: .

Rank of from largest to smallest .