Two polaroids have their polarizing directions parallel so that the intensity of a transmitted light is maximum. The angle through which either Polaroid must be turned if the intensity is to drop by one-half is

Question

Two polaroids have their polarizing directions parallel so that the intensity of a transmitted light is maximum. The angle through which either Polaroid must be turned if the intensity is to drop by one-half is (A) 135° (B) 90° (C) 120° (D) 180°

Tardigrade Admin · Accepted Answer

For I= (I0/2) and I=I0 cos2 θ=(I0/2) ∴ θ=45Â° Therefore the angle through which either polaroids turned is 135Â° (= 180Â° -45Â°)

Q. Two polaroids have their polarizing directions parallel so that the intensity of a transmitted light is maximum. The angle through which either Polaroid must be turned if the intensity is to drop by one-half is

$13 5^{°}$

$9 0^{°}$

$12 0^{°}$

$18 0^{°}$

For $I = \frac{I _{0}}{2} an d I = I_{0} co s^{2} θ = \frac{I _{0}}{2}$
$∴ θ = 4 5^{°}$
Therefore the angle through which either polaroids turned is $13 5^{°} (= 18 0^{°} - 4 5^{°})$

Q. Two polaroids have their polarizing directions parallel so that the intensity of a transmitted light is maximum. The angle through which either Polaroid must be turned if the intensity is to drop by one-half is

135°

90°

120°

180°

For I=2I0​​andI=I0​cos2θ=2I0​​ ∴θ=45° Therefore the angle through which either polaroids turned is 135°(=180°−45°)

$13 5^{°}$

$9 0^{°}$

$12 0^{°}$

$18 0^{°}$

For $I = \frac{I _{0}}{2} an d I = I_{0} co s^{2} θ = \frac{I _{0}}{2}$
$∴ θ = 4 5^{°}$
Therefore the angle through which either polaroids turned is $13 5^{°} (= 18 0^{°} - 4 5^{°})$