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

A transparent thin plate of a polaroid is placed on another similar plate such that the angle between their axes is $30^\circ$. The intensities of the emergent and the unpolarized incident light will be in the ratio of

  1. $1 : 4$
  2. $1 : 3$
  3. $3 : 4$
  4. $3 : 8$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $I _0$ be the intensity of unpolarized light, then intensity of light from first transparent thin plate of a polaroid is $I=\dfrac{I _0}2$


Now this light will pass through the second similar plate whose axis is inclined at an angle of $30^o$ to that of first plate.
According to Malus law, the intensity of emerging light is:-


$I'= Icos^2 30^o=\dfrac{I _0}2\bigg(\dfrac{\sqrt 3}{2}\bigg)^2=\dfrac 38 I _0$

$\therefore \dfrac{I'}{I _0}= \dfrac 38$

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

Unpolarised light of intensity $32\ W\ m^{-2}$ passes through three polarizers such that transmission axis of first is crossed with third. If intensity of emerging light is $2\ W\ m^{-2}$, what is the angle of transmission axis between the first two polarisers? 

  1. $30^{\circ}$
  2. $45^{\circ}$
  3. $22.5^{\circ}$
  4. $60^{\circ}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $\theta$ be angle between the axis of the first two polarisers then obviously $(90^o - \theta)$ is the angle between $2^{nd} \, and \, 3^{rd}$ polarisers. 
$\therefore I = \dfrac{I _0}{2} \, cos^2 \theta \, cos^2 (90^o - \theta)$ 

or $2 = \dfrac{32}{2} \, cos^2 \, \theta \, cos^2 \,  (90^o - \theta) = 16 cos^2 \theta \, sin^2 \theta$
$(2sin \theta \, cos \theta)^2 = \dfrac{1}{2} \, or \, sin^2 2 \theta = \dfrac{1}{\sqrt{2}}$
$\therefore 2 \theta = 45^o \, \ or \, \theta = 22.5^o$

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

On unpolarised beam of light is incident on a set of four polarising plates, such that each plate makes an angle of $\dfrac{\pi}{3}$ with preceding sheet. The light transmitted through the combination is:-

  1. $\dfrac{1}{128}$
  2. $\dfrac{1}{256}$
  3. $\dfrac{1}{64}$
  4. $\dfrac{1}{32}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

According to Malus's Law, when unpolarized light of intensity I_0 passes through the first polarizer, its intensity becomes I_0 / 2. Each subsequent polarizer reduces the intensity by cos^2(pi/3) = (1/4). With four polarizers in total, there are three successive transmissions, yielding (I_0 / 2) * (1/4)^3 = I_0 / 128.

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

Two polaroids are kept crossed to each other. Now one of the polaroids is rotated through an angle $45^0$. The percentage of incident light now transmitted through the system is: 

  1. 15%

  2. 25%

  3. 50%

  4. 60%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Initially crossed (90 degrees), I=0. Rotating one by 45 degrees means the angle between them is 45 degrees. Intensity I = (I0/2) * cos^2(45) = I0/4 = 25%.

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

A plane polarized light is incident normally on a tourmaline plate. Its $\vec { E } $ vectors make an angle of ${ 60 }^{ o }$ with the optic axis of the plate. Find the percentage difference between initial and final intensities.

  1. $25$%
  2. $50$%
  3. $75$%
  4. $90$%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the initial intensity of the light be  $I _o$.
Final intensity of the light   $I = I _o\cos^2\theta$
where  $\theta= 60^o$
$\implies \ I = I _o\times \cos^260^o = 0.25 I _o$
Percentage change in the initial and final intensities  $ = \dfrac{I _o - I}{I _o}\times 100 = \dfrac{I _o-0.25I _o}{I _o}\times 100 = 75$ %
So option C is correct.