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Wave Optics Test - 74

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Wave Optics Test - 74
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  • Question 1
    1 / -0
    In a Youngs double-slit experiment, let $$\beta$$ be the fringe width, and let $${l}_{0}$$ be the intensity at the central bright fringe. At a distance $$x$$ from the central bright fringe, the intensity will be:
    Solution
    $$\Delta =X\left( d/D \right)$$
    $$\therefore\quad$$ phase difference$$=\phi=\dfrac {2\pi}{\lambda}\Delta $$
    Let $$a=$$ amplitude at the screen due to each slit
    $$\therefore\quad {l}_{0}=k{(2a)}^{2}={4ka}^{2}$$, where $$k$$ is constant.
    For phase difference $$\phi$$, amplitude $$=A=2a \cos(\phi/2)$$.
    Intensity, $$l={ kA }^{ 2 }=k\left( { 4a }^{ 2 } \right) \cos ^{ 2 }{ \left( \phi /2 \right)  } { l }_{ 0 }\cos ^{ 2 }{ \left( \pi \Delta /\lambda  \right)  }$$
    $$={ l }_{ 0 }\cos ^{ 2 }{ \left( \dfrac { \pi  }{ \lambda  } .\dfrac { xd }{ D }  \right)  } ={ l }_{ 0 }\cos ^{ 2 }{ \left( \dfrac { \pi x }{ \beta  }  \right)  }.$$
    Hence $$(C)$$ is correct.
  • Question 2
    1 / -0
    In Young's double slit experiment, the slits are $$2$$ mm apart and are illuminated by photons of two wavelengths $$\lambda_1=12000\overset{o}{A}$$ and $$\lambda_2=10000\overset{o}{A}$$. At what minimum distance from the common central bright fringe on the screen $$2$$ m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?
    Solution
    $$\dfrac { { \lambda  }_{ 1 } }{ { \lambda  }_{ 2 } } =\dfrac { { h }_{ 1 } }{ { h }_{ 2 } } $$

    $$=\dfrac { 12000 }{ 10000 } =\dfrac { 6 }{ 5 } $$

    $$=\dfrac { { u }_{ 1 }{ \lambda  }_{ 1 }D }{ d } =\dfrac { 5\times 12000\times { 10 }^{ -10 }\times 2 }{ 2\times { 10 }^{ -3 } } $$

    $$=5\times 1.2\times { 10 }^{ 4 }\times { 10 }^{ -10 }\times { 10 }^{ 3 }=6mm$$
  • Question 3
    1 / -0
    A source and a listener arc moving towards each other with a speed $$(1/10)th$$ that of sound. If frequency of sound emitted by the source is $$f$$ than the frequency heard by listener will be
    Solution

  • Question 4
    1 / -0
    In $$YDSE$$, the central bright fringes is at mid-point of screen. On both side of mid-point fringes are obtained. Number of fringes on either side is $$n_1$$, with air between slits and screen. When water $$^a\mu_{W}=\dfrac {4}{3}$$ is filled $$'n_2'$$, number of fringes are obtained on either side, then $$n_{2}$$ is
    Solution

  • Question 5
    1 / -0
    Two coherent sources of wavelengths $$4.8\times {10}^{-7}m$$ produce a steady interference pattern
    The path difference corresponding to $$10th$$ order maximum will be
    Solution

  • Question 6
    1 / -0
    In a YDSE experiment if a slab whose refractive index can be varied is placed in front of one of the slits then the variation of resultant intensity at mid-point of screen with $$\mu(\mu\ge 1)$$ will be best represented by (assume  slits of equal width and there is not absorption by slab)
    Solution

  • Question 7
    1 / -0
    In a Young's double slit experiment , a slab of thickness $$1.2 \mu m$$ and refractive index $$1.5$$ is placed in front of one slit and another slab of thickness t and refractive index $$2.5$$ is placed in front of the second slit. If the position of the central fringe remains unaltered, then the thickness is

    Solution

  • Question 8
    1 / -0
    A source of sound emitting a note of frequency $$500\ Hz$$ moves towards an observer with a velocity $$v$$ equal to half the velocity of sound. If the observer moves away from the source with the same velocity, then the apparent frequency heard by the observer is
    Solution

  • Question 9
    1 / -0
    Consider that the screen starts moving towards right with speed $$v$$. At $$t=0$$, second maxima was lying at point $$P$$. At what time $$t$$, first minima will be lying at the point $$P$$?
    Solution

  • Question 10
    1 / -0
    A source of sound of frequency $$258\ Hz$$ is moving rapidly towards a wall with a velocity of $$5m/s$$. How many beats per second will be heard by the observer situated on the source itself if sound travels at a speed of $$330\ m/s$$
    Solution

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