Unpolarised light falls on two polarizing sheets placed one on top of the other. What must be the angle between the characteristic directions of the sheets if the intensity of the final transmitted light is one-third the maximum intensity of the first transmitted beam?

Question

Unpolarised light falls on two polarizing sheets placed one on top of the other. What must be the angle between the characteristic directions of the sheets if the intensity of the final transmitted light is one-third the maximum intensity of the first transmitted beam? (A) cos - 1((1/4)) (B) cos- 1((1/√3)) (C) cos- 1((1/√2)) (D) cos- 1((1/3))

Tardigrade Admin · Accepted Answer

I'=lcos2θ (I/3)=Icos2θ θ =cos- 1((1/√3))

Q. Unpolarised light falls on two polarizing sheets placed one on top of the other. What must be the angle between the characteristic directions of the sheets if the intensity of the final transmitted light is one-third the maximum intensity of the first transmitted beam?

$co s^{- 1} (\frac{1}{4})$

$co s^{- 1} (\frac{1}{3})$

$co s^{- 1} (\frac{1}{2})$

$co s^{- 1} (\frac{1}{3})$

$I^{^{'}} = l co s^{2} θ$
$\frac{I}{3} = I co s^{2} θ$
$θ = co s^{- 1} (\frac{1}{3})$

Q. Unpolarised light falls on two polarizing sheets placed one on top of the other. What must be the angle between the characteristic directions of the sheets if the intensity of the final transmitted light is one-third the maximum intensity of the first transmitted beam?

cos−1(41​)

cos−1(3​1​)

cos−1(2​1​)

cos−1(31​)

I′=lcos2θ 3I​=Icos2θ θ=cos−1(3​1​)

$co s^{- 1} (\frac{1}{4})$

$co s^{- 1} (\frac{1}{3})$

$co s^{- 1} (\frac{1}{2})$

$co s^{- 1} (\frac{1}{3})$

$I^{^{'}} = l co s^{2} θ$
$\frac{I}{3} = I co s^{2} θ$
$θ = co s^{- 1} (\frac{1}{3})$