Unpolarized light is incident upon three polarizers. The first polarizer, which

Question

Unpolarized light is incident upon three polarizers. The first polarizer, which is closest to the light source, has a vertical transmission axis. The intensity of the incident light is I_0 = 50.0 W/m^2. a) What is the intensity of the light after it has exited the first polarizer? b) If the intensity of the light after it has exited the second polarizer is 25.0% of the intensity of the incident unpolarized light (I_0), what angle does the transmission axis of the second polarizer make with the vertical? c) The transmission axis of the third polarizer makes an angle of 15.0 degree with the horizontal. What is the intensity of the light after it has exited the third polarizer? d) What would the intensity of the light exiting the third polarizer become if you were to remove the second polarizer? e) If you were to place a fourth polarizer with a horizontal transmission axis after the third one (assume the second polarizer is in place), would any light emerge from it? Justify your answer. f) How would your answers to questions a) - d) change if the incident light were polarized along a horizontal transmission axis? g) How would your answers to questions a) - d) change if the incident light were polarized along a vertical transmission axis?

Explanation / Answer

(A) As unpolarized light passes through the polarizer, its intensity gets halved.And light become polarized

I = Io / 2

I =50 / 2 = 25 W/m^2

(B) after that intensity,

I' = I ( cos(theta))^2

0.25 Io = 0.50 Io (cos(theta))^2

cos(theta) = 0.707

theta = 45 deg ......Ans

(c) as light passes through third polarizer,

theta= 90 - 45 - 15 = 30 deg

I = (0.25 x 50) (cos30)^2

I = 9.375 W /m^2

(d) if second polarizer removed than theta = 90 - 15 = 75 deg

I = (0.5 x 50) (cos75)^2 = 1.675 W/m^2

(e) now theta = 15 deg

so I = 9.375 cos(15)^2 = 8.75 W/m^2

hence light will emerge.

(For light not to emerge, two consecutive polarizer have to at right angle that is not case here, hence light will emerge)

(f) (a) I = 25 W/m^2

(b) theta = 90 -45 = 45 deg

(c) I = 9.375 W/m^2

(d) now theta = 15 deg

I = (0.5 x50) (cos15)^2 = 23.33 W/m^2

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

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