Q: Malus' law relates the intensity transmitted by an analyser at angle \theta to the incident plane-polarised intensity I_0 as:

I = I_0\cos^2\theta. Only the component of the electric field along the transmission axis passes, and intensity goes as the square of the amplitude.

Q: Brewster's law states that the refractive index of a medium equals:

The tangent of the polarising angle. \mu = \tan i_p. At this angle the reflected and refracted rays are perpendicular to one another.

Q: When unpolarised light of intensity I_0 passes through a single ideal polaroid, the transmitted intensity is:

\dfrac{I_0}{2}. Averaging \cos^2\theta over all the random orientations present in unpolarised light gives \dfrac{1}{2}.

Question 1

The phenomenon of polarisation establishes that light waves are:

Accepted Answer

Transverse. Only a transverse vibration can be restricted to a single plane; longitudinal waves such as sound cannot be polarised.

Question 2

Malus' law relates the intensity transmitted by an analyser at angle 	heta to the incident plane-polarised intensity I_0 as:

Accepted Answer

I = I_0\cos^2	heta. Only the component of the electric field along the transmission axis passes, and intensity goes as the square of the amplitude.

Question 3

Brewster's law states that the refractive index of a medium equals:

Accepted Answer

The tangent of the polarising angle. \mu = 	an i_p. At this angle the reflected and refracted rays are perpendicular to one another.

Question 4

At Brewster's angle of incidence, the reflected ray is:

Accepted Answer

Completely plane-polarised perpendicular to the plane of incidence. The component vibrating in the plane of incidence cannot be radiated along the reflected direction, leaving only the perpendicular component.

Question 5

Two polaroids with their transmission axes perpendicular to each other transmit:

Accepted Answer

No light at all. By Malus' law I = I_0\cos^2 90^\circ = 0; crossed polaroids extinguish the beam.

Question 6

Sound waves cannot be polarised because they are:

Accepted Answer

Longitudinal waves. In a longitudinal wave the vibration is already along the direction of propagation, so there is no transverse plane to select.

Question 7

Polaroid sunglasses reduce glare from a road surface because the reflected light is:

Accepted Answer

Largely polarised horizontally, which the polaroid blocks. Reflection from a horizontal surface polarises light mainly in the horizontal plane, so a vertically oriented polaroid removes most of it.

Question 8

When unpolarised light of intensity I_0 passes through a single ideal polaroid, the transmitted intensity is:

Accepted Answer

\dfrac{I_0}{2}. Averaging \cos^2	heta over all the random orientations present in unpolarised light gives \dfrac{1}{2}.

Question 9

At the polarising angle, the angle between the reflected and the refracted rays is:

Accepted Answer

90^\circ. This perpendicularity is the geometrical content of Brewster's law and is what makes the reflected beam fully polarised.

Question 10

Unpolarised light of intensity I_0 passes through an ideal polaroid. The transmitted intensity is:

Accepted Answer

\dfrac{I_0}{2}. A polaroid transmits on average half of unpolarised incident light.

Question 11

Plane-polarised light of intensity I falls on an analyser whose axis makes 60^\circ with the polariser. The transmitted intensity is:

Accepted Answer

\dfrac{I}{4}. By Malus' law I' = I\cos^2 60^\circ = I 	imes \left(\dfrac{1}{2}ight)^2 = \dfrac{I}{4}.

Question 12

Plane-polarised light of intensity I falls on an analyser at 30^\circ to the polariser. The transmitted intensity is:

Accepted Answer

\dfrac{3I}{4}. I' = I\cos^2 30^\circ = I 	imes \dfrac{3}{4}.

Question 13

Plane-polarised light of intensity I falls on an analyser at 45^\circ to the polariser. The transmitted intensity is:

Accepted Answer

\dfrac{I}{2}. I' = I\cos^2 45^\circ = I 	imes \dfrac{1}{2}.

Question 14

Two polaroids are crossed at 90^\circ. A third polaroid is inserted between them at 45^\circ to the first. If unpolarised light of intensity I_0 is incident, the final transmitted intensity is:

Accepted Answer

\dfrac{I_0}{8}. After the first polaroid \dfrac{I_0}{2}; after the 45^\circ sheet \dfrac{I_0}{2}\cos^2 45^\circ = \dfrac{I_0}{4}; after the last, again at 45^\circ, \dfrac{I_0}{8}.

Question 15

The refractive index of a medium is \sqrt{3}. Its polarising angle is:

Accepted Answer

60^\circ. 	an i_p = \mu = \sqrt{3} \Rightarrow i_p = 60^\circ.

Question 16

The polarising angle for a medium of refractive index 1.5 is approximately:

Accepted Answer

56.3^\circ. i_p = 	an^{-1}(1.5) \approx 56.3^\circ.

Question 17

The polarising angle for water of refractive index \dfrac{4}{3} is approximately:

Accepted Answer

53.1^\circ. i_p = 	an^{-1}\left(\dfrac{4}{3}ight) = 	an^{-1}(1.333) \approx 53.1^\circ.

Question 18

Unpolarised light of intensity I_0 passes through two polaroids whose axes are at 60^\circ. The final transmitted intensity is:

Accepted Answer

\dfrac{I_0}{8}. The first polaroid gives \dfrac{I_0}{2}; the second transmits \cos^2 60^\circ = \dfrac{1}{4} of that, leaving \dfrac{I_0}{8}.

Question 19

Light is incident on a medium at the polarising angle of 60^\circ. The angle of refraction is:

Accepted Answer

30^\circ. At the polarising angle the reflected and refracted rays are perpendicular, so r = 90^\circ - i_p = 30^\circ.

Question 20

Plane-polarised light falls on an analyser whose axis is parallel to the plane of polarisation. The transmitted intensity is:

Accepted Answer

Equal to the incident intensity. With 	heta = 0, Malus' law gives I = I_0\cos^2 0 = I_0.

Polarisation of Light — MH-CET Physics MCQs with Solutions

Questions with solutions