Two linear polarizing filters are placed one behind the other so that their transmission directions form an angle of 45 ° . A beam of unpolarized light of intensity I 0 is directed at the two filters. What fraction of the light will pass through both filters?

The transmitted intensity will be the quarter of the incident intensity of light, (I = 0.25 I₀), when two linear polarizing filters are placed one behind the other forming the angle of 45°.

Explanation:

Polarizers are used for constraining the direction of propagation of light waves.  So depending upon the angle of polarizer, the transmitted rays will follow that angle only and the remaining lights waves at different angles will be reflected back. So basically it works like a filter or optical filters.

So, the transmitted intensity will be less compared to the original intensity of light as the beam will get confined in a particular angle.

The relation between transmitted intensity and incident intensity is given by Malus's law.

Here, I is the transmitted intensity, I₀ is the incident intensity and α is the angle of the polarizer.

In the present case, two polarizers are used with same angle, so the first one termed as polarizer, while the second one will be termed as analyzer. The angle which is used to polarize the intensity of the incident beam is the angle of 45°.

Then, according to Malus's law,

I = I₀ cos² 45 = (I₀/2)×(1/√2)² = I₀/4 =

So, the transmitted intensity will be the quarter of the incident intensity of light, (I = 0.25 I₀), when two linear polarizing filters are placed one behind the other forming the angle of 45°.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s