? webassign.net Search 0.5/1 points | Previous Answers In the figure below, a li

Question

? webassign.net Search 0.5/1 points | Previous Answers In the figure below, a light is incident at angle 0, 529 on a series of five transparent layers with paraliei boundaries. For layers 1 and 3, L1 -17 um, L3 26 um, n 1.80, and n3 1.59. My Not Air Air 8, (a) At what angle does the light emerge back into air at the right? 52 (b) How long does the light take to travel through layer 3? 0.647 x ps nsparent layer .50 (a) At what hrough layer 0-1 points My Two waves of light in air, of wavelength A- 610.0 nm, are initially in phase. They then travel through plastic layers as shown in Figure 35-36, with Li-4.00 ?m, L2-3.50 ?m, ni-1.30, and n2-1.60. nsparent 46. (b) How tin 0.12 pm

Explanation / Answer

According to snell's law of refraction:

?1sin?1 = ?2sin?2

Where ?1 is the refractive index of the incident medium

?2 is the refractive index of the refraction medium

?1 is the incident angle

?2 is the refraction angle

For the first boundary

1*sin52 = 1.80*sin?

? = 26o

For the second boundary

1.80*sin26 = n*sin?

n*sin? = 0.788

For the third boundary

0.788 = 1.59*sin?

? = 29.7o

For the fourth boundary

1.59*sin29.7 = n*sin?

n*sin? = 0.788

For the fifth boundary

0.788 = n*sin?

n*sin? = 0.788

For the fourth boundary

0.788 = sin?

? = 520

The time taken for the light to travel through the third layer of thickness 26 x 10-6 m is

T = distance/velocity

T = ?t/c

T = 1.59 x 26 x 10-6 / 3 x 108

T = 1.378 x 10^-13 s

T = 13.78 x 106-14 s

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.