Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

Multiple choice multiple-slit diffraction the principle of superposition of waves superposition of waves oscillations and waves physics

In an interference experiment, phase difference for points where the intensity is minimum is (n = 1, 2, 3 ...)

  1. $n \pi$

  2. $(n\, +\, 1) \pi$

  3. $(2n\, +\, 1) \pi$

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

Intensity at a point due to interference of beams of intensities $I _1,I _2$ with a phase difference $\phi$ between them=$I _1+I _2+2\sqrt{I _1I _2}cos\phi$

The value of the resultant intensity is minimum for $cos\phi=-1$
$\implies \phi=(2n+1)\pi$ for positive integer values of $n$

Multiple choice multiple-slit diffraction the principle of superposition of waves superposition of waves oscillations and waves physics

Two identical light waves, propagating in the same direction, have a phase difference $\delta $. After they superpose the intensity of the resulting wave will be proportional to

  1. $\cos { \delta } $

  2. $\cos { \left( \dfrac { \delta }{ 2 } \right) } $

  3. $\cos ^{ 2 }{ \left( \dfrac { \delta }{ 2 } \right) } $

  4. $\cos ^{ 2 }{ \delta } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Maximum intensity,
     $I=4{ I } _{ 0 }\cos ^{ 2 }{ \left( \dfrac { \delta  }{ 2 }  \right)  } $
$\Rightarrow I\propto \cos ^{ 2 }{ \left( \dfrac { \delta  }{ 2 }  \right)  } $

Multiple choice multiple-slit diffraction the principle of superposition of waves superposition of waves oscillations and waves physics

In the Young's double slit experiment, the resultant intensity at a point on the screen is 75% of the maximum intensity of the bright fringe. Then the phase difference between the two interfering rays at that point is 

  1. $\frac {\pi}{6}$

  2. $\frac {\pi}{4}$

  3. $\frac {\pi}{3}$

  4. $\frac {\pi}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The correct answer is option(C).

We know, The resultant intensity,
$I _R=4I _{max}cos^2\left( \frac \phi 2\right)$
$or, cos^2\left( \frac \phi 2\right)=\frac {I _R}{4I _max}=\frac {0.75}4=0.1875$
$\Rightarrow cos\left(\frac \phi 2\right)=\sqrt {0.1875}=0.5$
$\Rightarrow \frac \phi 2 = cos _{-1}(0.5)=\frac \pi 3$

Multiple choice multiple-slit diffraction the principle of superposition of waves superposition of waves oscillations and waves physics

Two beams of light having intensities I and 4 I interface to produce a fringe pattern on a screen. The phase difference between the beams is $\dfrac { \pi  }{ 2 }$ at point A and $\pi$ at point B, Then the difference between the resultant intensities at A and B is

  1. 2 I

  2. 4 I

  3. 5 I

  4. 7I
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Resultant intensity at a point, $I _R= I _1+ I _2 +2\sqrt{I _1I _2} cos \phi$

where $I _1$ and $I _2$ be the intensities of two sources and $\phi$ be the phase differences of the sources at that point.
Here, $I _1= I$ and $I _2=4I$

At point $A$, $\phi= \dfrac{\pi}2$
So,Resulant intensity at point $A$,   $I _A= 5I+ 2\sqrt{4I^2}cos \dfrac{\pi}2= 5I$   

At point $B$, $\phi=\pi$

So,Resulant intensity at point $B$,   $I _B= 5I+ 2\sqrt{4I^2}cos\pi= 5I-4I= I$

Hence, Required difference between the intensities at $A$ and $B= I _A-I _B= 5I-I= 4I$

Multiple choice multiple-slit diffraction the principle of superposition of waves superposition of waves oscillations and waves physics

In Young's double slit experiment, the two slits act as coherent sources of waves of equal amplitude $A$ and wavelength $\lambda$. In another experiment with the same arrangement, the two slits are made to act as incoherent sources of waves of same amplitude and wavelength. If the intensity at the middle point of the screen in the first case is $I _1$ and in the second case is $I _2$, then the ratio $I _1/I _2$ is: 

  1. $0.5$

  2. $4$

  3. $2$

  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice multiple-slit diffraction the principle of superposition of waves superposition of waves oscillations and waves physics

Two beams of light having intensities $I$ and $4I$ interfere to produce a fringe pattern on a screen.If the phase difference between the beams is $\dfrac{\pi }{2}$ at point A and $\pi$ at point B then the difference between the resultant intensities at A and B is 

  1. 4I

  2. 2I

  3. 5I

  4. 7I
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The resultant intensity is given by:
$I={I} _{1}+{I} _{2}+2\sqrt{{I} _{1}{I} _{2}}\cos{\phi}$
Thus difference is given by:
${I} _{A}-{I} _{B}=2\sqrt{{I} _{1}{I} _{2}}(\cos{\dfrac{\pi}{2}}-\cos{\pi})=2\sqrt{4{I}^{2}}\times1=4I$

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

In forced oscillation of a particle the amplitude is maximum for a frequency $\omega _1$ of force, while the energy is maximum for a frequency $\omega _2$ of the force, then:

  1. $\omega _1= \omega _2$

  2. $\omega _1> \omega _2$

  3. $\omega _1 < \omega _2$ when damping is small and $\omega _1> \omega _2$ when damping is large

  4. $\omega _1< \omega _2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the amplitude of oscillation and energy to be maximum, the frequency of force must be equal to the initial frequency and this is only possible in resonance. In resonance state $ \omega _1 = \omega _2$.

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

If density (D) acceleration (a) and force (F) are taken as basic quantities,then Time period has dimensions

  1. $\dfrac {1} {6}$ in F

  2. $-\dfrac {1} {6}$ in F

  3. $-\dfrac {2} {3}$ in F

  4. All the above are true
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The density has dimension ML-3

so take that as D

the acceleration has dimension LT-2

take it as A

the force has dimension MLT-2

Take it as F

so time period has dimension as D-1/6 A -2/3 F1/6

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

The potential energy of a particle of mass $1\ kg$ in motion along the $x-$axis is given by: $U=4(1-\cos 2x)\ l$, where $x$ is in metres. The period of small oscillations (in sec) is: 

  1. $2\pi$

  2. $\pi$

  3. $\dfrac{\pi}{2}$

  4. $\sqrt {2}\pi$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For small oscillations, U = 4(1 - cos(2x)) approx 4(1 - (1 - (2x)^2/2)) = 4(2x^2) = 8x^2. The force F = -dU/dx = -16x. Comparing to F = -kx, k = 16. Period T = 2*pi * sqrt(m/k) = 2*pi * sqrt(1/16) = 2*pi * 1/4 = pi/2.