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 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  $\dfrac{I _{unpolarized}}{2}=16Wm^{-2}$

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 $I$. The value of the $I$ is 

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

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

When the unpolarized light falls on the first tourmaline crystal, the intensity of the light halves and becomes polarized.

Thus $I _1=\dfrac{I _0}{2}$
When this polarized light falls on the next tourmaline crystal at an angle $\theta$, the intensity of light becomes,
$I _2=I _1cos^2\theta=I _1 cos^20^{\circ}$
$=\dfrac{I _0}{2}$

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 $I$. Flux of energy of the incident ray is $10^{-3}W$, the percentage of incident light transmitted by the second polarizered will be____

  1. $12.5\%$
  2. $25\%$
  3. $37.5\%$
  4. $50\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $I$. The intensity of the emergent beam, if flux of energy of the incident ray is $10^{-3}W$, will be (in $W/m^2$)

  1. $\displaystyle\frac{I}{3}$
  2. $\displaystyle\frac{2I}{3}$
  3. $\displaystyle\frac{4I}{3}$
  4. $\displaystyle\frac{5I}{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice physics oscillations and waves polarization of light polarisation of light polarisation

A Plane polarized light is incidents on an analyzer. The intensity then becomes three-fourth. The angle of the axis of the analyzer with the beam is

  1. $30^0$
  2. $45^0$
  3. $60^0$

  4. zero
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Intensity of light after passing through an analyzer which is at an angle $\theta$ with the beam,

$I=I _0\cos^2\theta$
$\implies \dfrac{3}{4}I _0=I _0\cos^2\theta$
$\cos\theta=\dfrac{\sqrt{3}}{2}$
$\implies \theta=30^{\circ}$