Tag: oscillations and waves

Quizzes (22) Questions (115)

Questions Related to oscillations and waves

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 $
$I _{t}=\dfrac{I _{0}}{2}\times \dfrac{1}{2}=\dfrac{I _{0}}{4}$

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

Two Polaroids $P _1$ and $P _2$ are placed with their axis perpendicular to each other. Unpolarized light $l _0$ is incident on $P _1$. A third polaroid $P _3$ is kept in between $P _1$ and $P _2$ such that its axis makes an angle $45^{\circ}$ with that of $P _1$. The intensity of transmitted light through $P _2$ is 

  1. $\frac {I _0 }{2 }$
  2. $\frac {I _0 }{4 }$
  3. $\frac {I _0 }{8 }$
  4. $\frac {I _0 }{16 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$I=I _0 .\cos^2 \theta$
$\theta =$ angle made by $E$ vector with transmission axis.wherein
$I=$ Intensity of transmitted light after polarisation.
$I _0=$Intensity of incident light.

Intensity of light after crossing $P _1=\dfrac {I _0}{2}$
Intensity of light after crossing $P _3=\dfrac {I _0}{2}.\cos^2 45^o =\dfrac {I _0}{4}$
Intensity of light after crossing $P _2=\dfrac {I _0}{4}.\cos^2 45^o$
$I=\dfrac {I _0}{8}$
Multiple choice physics oscillations and waves polarization of light polarisation of light polarisation

Three polaroides are placed one above other, such that the first and the last polaroids are crossed with each other. If the angle between the transmission axis of the first two polaroids is $45$, then what is the percentage of incident light transmitted through the combination of three polaroids?

  1. 0%

  2. 12.50%

  3. 50%

  4. 100%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

After the first polarizer, intensity is I0/2. After the second (at 45 degrees), I2 = (I0/2)cos^2(45) = I0/4. After the third (crossed with the first, so 45 degrees to the second), I3 = I2*cos^2(45) = (I0/4)(1/2) = I0/8 = 12.5%.

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

For a given medium, the polarising angle is $60^o$. What will be the critical angle for this medium?

  1. $i = 35^o16'$
  2. $i = 45^o16'$
  3. $i = 55^o16'$
  4. $i = 65^o16'$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Brewster's Law states tan(ip) = n. Given ip = 60 degrees, n = tan(60) = sqrt(3). The critical angle c is given by sin(c) = 1/n = 1/sqrt(3). c = arcsin(1/sqrt(3)) is approximately 35.26 degrees, which is 35 degrees 16 minutes.

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

A light has amplitude A and angle between analyzer and polarizer is.$60 ^ { \circ }$ Light is transmitted by analyzer has amplitude. 

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

Intensity I = k*A^2. After the polarizer, the amplitude is A/sqrt(2). After the analyzer at 60 degrees, the amplitude becomes (A/sqrt(2))cos(60) = (A/sqrt(2))(1/2) = A/(2*sqrt(2)).