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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)).

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

The polaroids are placed in the path of unpolarized beam of intensity $I _{0}$ such that no light is emitted from the second polaroid. If a third polaroid whose polarization axis makes an angle $\theta$ with the polarization axis of first polaroid, is placed between these polariods then the intensity of light emerging from the last polaroid will be

  1. $\left (\dfrac {I _{0}}{8}\right )\sin^{2} 2\theta$
  2. $\left (\dfrac {I _{0}}{4}\right )\sin^{2} 2\theta$
  3. $\left (\dfrac {I _{0}}{2}\right )\sin^{2} 2\theta$
  4. $I _{0}\cos^{4}\theta$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Intensity after the first polarizer is I0/2. The second polarizer is at 90 degrees to the first. The third polarizer is at angle theta to the first (so 90-theta to the second). The intensity after the third is (I0/2)cos^2(theta)*cos^2(90-theta) = (I0/2)*cos^2(theta)*sin^2(theta) = (I0/2)(sin(2*theta)/2)^2 = (I0/8)*sin^2(2*theta).

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

A Polaroid examines two adjacent plane polarised beams $A$ and $B$ whose planes of polarisation are mutually perpendicular. In the first position of the analyser, beam $B$ shows zero intensity. From this position a rotation of $30^{o}$ shows that the two beams have same intensity. The ratio of intensities of the two beams $I _{A}$ and $I _{B}$ will be

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

Let the analyzer be at angle alpha. Intensity I = I_A*cos^2(alpha) + I_B*cos^2(alpha+90) = I_A*cos^2(alpha) + I_B*sin^2(alpha). At alpha = 30 degrees, I_A*cos^2(30) = I_B*sin^2(30). I_A*(3/4) = I_B*(1/4), so I_A/I_B = 1/3.

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

In, the visible region of the spectrum the rotation of the plane of polarization is given by $\displaystyle\theta=a+\frac{b}{\lambda^2}$. The optical rotation produced by a particular material is found to be $30^0$ per $mm$ at $\lambda=5000A^o$ and $50^0$ per $mm$ at $\lambda=4000A^o$. The value of constant $a$ will be

  1. $\displaystyle +\frac{50^0}{9}$ per $mm$
  2. $\displaystyle -\frac{50^0}{9}$ per $mm$
  3. $\displaystyle +\frac{9^0}{50}$ per $mm$
  4. $\displaystyle -\frac{9^0}{50}$ per $mm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using theta = a + b/lambda^2: 30 = a + b/(5000^2) and 50 = a + b/(4000^2). Subtracting the equations: 20 = b*(1/16*10^6 - 1/25*10^6) = b*(9/400*10^6). b = 20 * 400*10^6 / 9. Substituting back: a = 30 - (20 * 400*10^6 / 9) / 25*10^6 = 30 - 320/9 = (270-320)/9 = -50/9.

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

An unpolarized beam of intensity $2a^2$ passes through a thin Polaroid. Assuming zero absorption in the Polaroid, the intensity of emergent planes polarized light will be  

  1. $2a^2$
  2. $a^2$
  3. $\displaystyle\sqrt2a$
  4. $\displaystyle\frac{a^2}{\sqrt2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 initial unpolarized intensity is $2a^{2}$ 
the intensity of light transmitted by the first polarizered will be  $\dfrac{I _{unpolarized}}{2}=a^{2}$
option $B$ is correct 

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

A beam of unpolarized light is passed first through a tourmaline crystal $A$ and then through another tourmaline crystal $B$ oriented so that its principal plane is parallel to that of $A$. The intensity of final emergent light is $I$. If $A$ is rotated by $45^0$ on a plane, perpendicular to the direction of incident ray, then intensity of emergent light will be

  1. $\displaystyle\frac{I}{8}$
  2. $\displaystyle\frac{I}{4}$
  3. $\displaystyle\frac{I}{2}$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$I _{I}=Icos^{2} \theta =Icos^{2}45=\dfrac{I}{2}$
option $C$ is correct 

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

Unpolarized light of intensity $32Wm^{-2}$ passes through three polarizes such that the transmission axis of the last polarizers is crossed with that of the first. The intensity of final emerging light is $3Wm^{-2}$.The intensity of light transmitted by the first polarizered will be 

  1. $32Wm^{-2}$
  2. $16Wm^{-2}$
  3. $8Wm^{-2}$
  4. $4Wm^{-2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 initial unpolarized intensity is $32Wm^{-2}$
the intensity of light transmitted by the first polarizered will be  $\dfrac{I _{unpolarized}}{2}=16Wm^{-2}$