Tag: physics

Questions Related to physics

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

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

  1. $n \pi$

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

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

  4. zero

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Correct Option: C
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 $n$

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

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

  1. $\cos { \delta } $

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

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

  4. $\cos ^{ 2 }{ \delta } $

Show answer
Correct Option: C
Explanation:

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

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 

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

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

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

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

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

Show answer
Correct Option: C
Explanation:

The correct answer is option(C).

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

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

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

  1. 2 I

  2. 4 I

  3. 5 I

  4. 7I

Show answer
Correct Option: B
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 $B= I _A-I _B= 5I-I= 4I$

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: 

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

  1. $0.5$

  2. $4$

  3. $2$

  4. $1$

Show answer
Correct Option: A

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 

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

  1. 4I

  2. 2I

  3. 5I

  4. 7I

Show answer
Correct Option: A
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:
${I} _{A}-{I} _{B}=2\sqrt{{I} _{1}{I} _{2}}(\cos{\dfrac{\pi}{2}}-\cos{\pi})=2\sqrt{4{I}^{2}}\times1=4I$

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:

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

  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. $\omega _1< \omega _2$

Show answer
Correct Option: A
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$.

A weightless spring has a force constant $k$ oscillates  with frequency $f$ when a mass $m$ is suspended from it. The spring is cut into three equal parts and a mass $3\ m$ is suspended from it. The frequency of oscillation of one part  will now becomes

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

  1. $f$

  2. $2\ f$

  3. $f/3$

  4. $3\ f$

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Correct Option: C

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

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

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

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

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

  4. All the above are true

Show answer
Correct Option: A
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

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: 

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

  1. $2\pi$

  2. $\pi$

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

  4. $\sqrt {2}\pi$

Show answer
Correct Option: C