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Questions Related to physics

Multiple choice physics wave optics polarization of light polarisation of light polarisation

A point source of monochromatic light is situated at the centre of a circle, what is the phase difference between the light waves passing through the end points of any diameter

  1. $\dfrac{\pi}{2}$
  2. $\pi$
  3. $\dfrac{3\pi}{2}$
  4. $zero$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A point source emits spherical waves. Any points at the same distance from the source, such as the endpoints of a diameter, lie on the same wavefront and thus have zero phase difference.

Multiple choice physics wave optics polarization of light polarisation of light polarisation

An unpolarised light of intensity $32  \mathrm{W} / \mathrm{m}^{2}  $ passes through three polarisers, such that the transmission axis of last polarizer is perpendicular with the first. If the intensity of emergent light is $3  \mathrm{Wh}  $ Im $ ^{2} $ then the angle between the transmission axes of the first two polarisers is:

  1. $ 30^{\circ} $
  2. $ 19^{\circ} $
  3. $ 45^{\circ} $
  4. $ 90^{\circ} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Malus's Law, the intensity after the first polarizer is I1 = I0/2 = 16. After the second, I2 = 16*cos^2(theta). After the third (perpendicular to the first), I3 = I2*cos^2(90-theta) = 16*cos^2(theta)sin^2(theta) = 16(sin(2*theta)/2)^2 = 4*sin^2(2*theta). Setting 4*sin^2(2*theta) = 3 gives sin^2(2*theta) = 3/4, so sin(2*theta) = sqrt(3)/2, meaning 2*theta = 60 degrees, or theta = 30 degrees.

Multiple choice physics oscillations and waves polarization of light polarisation of light polarisation

Two polaroids are oriented with their transmision axes making angle of $30^{\circ}$ with each other. The fraction of indicent unpolarised light is transmitted.

  1. $37\%$
  2. $37.5\%$
  3. $3.36\%$
  4. $36.5\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The intensity after the first polarizer is I0/2. The intensity after the second polarizer is I = (I0/2)cos^2(30). Since cos(30) = sqrt(3)/2, cos^2(30) = 3/4. Thus, I = (I0/2)(3/4) = 3/8 * I0 = 0.375 * I0, which is 37.5%.

Multiple choice physics oscillations and waves polarization of light polarisation of light polarisation

The axes of the polarizer and analyzer are inclined to each other at an angle of $60^{o}$. If the amplitude of polarized light emerging through the analyzer is $A$, the amplitude of unpolarized light incident on the polarizer is

  1. $A/2$
  2. $A$
  3. $2A$
  4. $2\sqrt{2}\ A$.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let I0 be the initial intensity. After the polarizer, intensity is I1 = I0/2. After the analyzer, I2 = I1*cos^2(60) = (I0/2)*(1/4) = I0/8. Since intensity is proportional to the square of amplitude (I = k*A^2), A_emergent^2 = (A_incident^2)/8. Thus, A_incident = A*sqrt(8) = 2*sqrt(2)*A.

Multiple choice physics oscillations and waves polarization of light polarisation of light polarisation

Two polorides are placed having their transmission axis at an angle of $45^0$. If unpolarised light is incident on first polorid acting as polarizer then, calculate intensity of emergent light from second polariser:-

  1. $I _0$
  2. $\frac{I _0}{4}$
  3. $\frac{I _0}{2}$
  4. $2I _0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Intensity after the first polarizer is I1 = I0/2. Applying Malus's Law for the second polarizer at 45 degrees: I2 = I1*cos^2(45) = (I0/2)*(1/2) = I0/4.

Multiple choice physics oscillations and waves polarization of light polarisation of light polarisation

Plane polarized light is passed through a Polaroid. Now the Polaroid is given one complete rotation about the direction of light propagation. When viewed through another Polaroid (analyser), one of the following is observed:

  1. The intensity of light gradually decreases to zero and then remains zero

  2. The intensity of light becomes twice maximum and twice zero

  3. The intensity of light becomes maximum and stays maximum

  4. The intensity of light does not change
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The Intensity of light coming out of a polaroid is
given by $I=I _{0}  cos^{2}\theta $
when $\theta $ is changed to $\theta +2\pi $  by rotation
$cos^{2}\theta$ becomes 1 twice at $\pi $ and $2\pi $ and
o twice at $\pi /2$ and $3\pi /2$
Thus option B is correct.

Multiple choice physics oscillations and waves polarization of light polarisation of light polarisation

When an unpolarised light is polarized, then the intensity of light of the polarized wave :

  1. remains the same

  2. gets doubled

  3. gets halved

  4. depends on the colour of the light.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In an unpolarised light the light has Intensity distributed in all polarizing directions.
When it is polarized half the Intensity is polarized in a particular direction and the other half in a perpendicular direction is not transmitted.

Multiple choice physics oscillations and waves polarization of light polarisation of light polarisation

Unpolarized light of intensity $I _{0}$ is incident on a polarizer and the emerging light strikes a second polarizing filter with its axis at 45$^{\circ}$ to that of the first. The intensity of the emerging beam :

  1. $\dfrac{I _{}o}{2}$
  2. $\dfrac{I _{}o}{4}$
  3. $I _{o}$
  4. $\dfrac{I _{}o}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The unpolarized light of Intensity $I _{0}$ is incident on a polarizer then a polarized light at intensity $I _{0}/2$ comes out.
Intensity of transmitted beam
$I _{t}=I _{0}/2 cos^{2} 45$ from the formula $I=I _{0} cos^{2} \theta $
$I _{t}=\dfrac{I _{0}}{2}\times \dfrac{1}{2}=\dfrac{I _{0}}{4}$