QUESTION 5 How much of the incident light is transmitted through the third polar

Question

QUESTION 5

How much of the incident light is transmitted through the third polarizer?

Link to photo: http://tinypic.com/view.php?pic=2zpv5ax&s=9

All

Some

None

Not enough information

QUESTION 6

In the polarizer question above, what is the polarization of the light transmitted through the third polarizer?

Vertical

Horizontal

45 degrees

None of these, since the light doesn't make it through the 3rd polarizer

QUESTION 7

Unpolarized light with brightness I0 encounters a polarizer which is oriented approximately vertically, letting through only vertically polarized light. This is then passed through a second polarizer oriented at an angle of to the first polarizer. The person on the right receives only about 1/3 of the original light (i.e., he observes an intensity of I0 /3). What is the approximate angle, , of the second polarizer? (Hint:use Malus's Law).

link to photo: http://tinypic.com/view.php?pic=1mbg6&s=9

57

17

70

0

All

Some

None

Not enough information

Explanation / Answer

Question 5:

By Malus’s law,

I1=I0/2 ---------------(1)

I2=I1*cos^2(45) ------------(2)

I3=I2*cos^2(45) ------------(2)

From (1), (2) and (3)

I3=I0/2*cos^2(45)*cos^2(45)= I0/8

Thus some of the incident light is transmitted through the third polarizer.

Question 6:

Angle of the polarization of the light transmitted through the third polarizer = Angle of the third polarizer = 0 deg wih respect to the horizontal.

Hence the polarization of the light transmitted through the third polarizer is horizontal.

Question 7:

By Malus’s law,

I1=I0/2 ---------------(1)

I2=I1*cos^2() ------------(2)

From (1) and (2)

I2=I0/2*cos^2()-----------(3)

I2= I0/3

From (3)

I0/3 = I0/2*cos^2()

1/3 = 1/2*cos^2()

= cos^-1[sqrt(2/3)] = 35.26 deg

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

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