Question

A beam of unpolarized light with intensity 170 W/m^2 is incident on a pair of ideal crossed (perpendicular) polarizers. You insert a third ideal polarizer between the two polarizers with its polarizing axis at 45∘ to the others. Determine the intensity of the emerging light after you insert the third polarizer. Express your answer with the appropriate units.

On February 23, 1987 astronomers noticed that a relatively faint star in the Tarantula Nebula in the Large Magellanic Cloud suddenly became so bright that it could be seen with the naked eye. Astronomers suspected that the star exploded as a supernova. Late-stage stars eject material that forms a ring before they explode as supernovas. The ring is at first invisible. However, following the explosion the light from the supernova reaches the ring. Astronomers observed the ring exactly one year after the phenomenon itself occurred (Figure 1). The angular size of the ring is 0.81 arcseconds.

Use geometry and the speed of light to estimate the distance to the supernova.

Answer 1

Solution

1)

The intensity of unpoloried light after passing through the first polorizer will be

I₁ = I_o/2 = 170 /2 = 85 W/m²

The intensity of light from the second polarizer

I₂ = I₁cos²45 = 85cos²45 = 42.5 W/m²

The intensity of light from the third polarizer

I₃ = I₂cos²45 = 42.5cos²45 = 21.25 W/m²

2)

The light from the super nova which is at the center of the ring reaches the ring after one year.

The ring is located one light year from the center. So

The radius of the ring

R = 1 light year = 9.46 x 10¹⁵ m

θ = 0.81 arcseconds. = 0.81 / 3600 degrees

θ/2 = 0.81 / 2 x 3600 = 0.0001125^o

From the diagram

Tan(θ/2) = R/d

Tan(0.0001125^o) = 9.46 x 10¹⁵/d

d = 9.46 x 10¹⁵ / 1.96 x 10^-6

d = 4.827 x 10²¹ m

Alternate method

R = 1 light year = 9.46 x 10¹⁵ m

θ = 0.81 arcseconds. = 0.81 x π/64800 radians

θ/2 = 0.81 x π / 2 x 648000 = 1.96 x 10^-6 radians

From the diagram

Tan(θ/2) = R/d

For extremely small angles θ, the tanθ = θ in radians

Tan(θ/2) = θ/2

1.96 x 10^-6 = 9.46 x 10¹⁵/d

d = 9.46 x 10¹⁵ / 1.96 x 10^-6

d = 4.827 x 10²¹ m