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Wave Optics Test - 33

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Wave Optics Test - 33
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  • Question 1
    1 / -0
    An astronomical telescope has a large aperture to
    Solution

  • Question 2
    1 / -0
    A very thin film in reflected white light appears
    Solution
    When the film is thin, t $$\rightarrow$$ 0, path diff. = $$\lambda$$\2. Therefore, in reflected light, the film appears black.
  • Question 3
    1 / -0
    The apparent wavelength of the light from star moving away from the earth is 0.2% more than its actual wavelength.Then the velocity of the star is 
    Solution
    Apparent frequency for a light source moving away with velocity $$u$$, can be given by:
    $$\nu' = \dfrac{c}{c+u}\nu_0= \dfrac{c}{\lambda'}$$

    Thus, $$\lambda'= \dfrac{c+u}{\nu_0}$$

    Now, $$\Delta \lambda = \lambda'-\lambda=\dfrac{u}{\nu_0}=\dfrac{u}{c}\lambda_0$$

    $$\Rightarrow \dfrac{\Delta \lambda}{\lambda_0}\times 100\% =\dfrac{u}{c}\times 100$$

    $$\Rightarrow 0.2 =\dfrac{u}{c}\times 100 $$

    $$\Rightarrow u= \dfrac{3\times 10^8 \times 0.2}{100}=6 \times 10^5$$ $$m/s$$
  • Question 4
    1 / -0
    Astronauts look down on earth surface from a space ship parked at an altitude of $$500$$ km. They can resolve object of the earth of the size (It can be assumed that the pupils diameter is 5 mm and wavelength of light is $$500$$ nm)
    Solution
    Resolving power of eye $$= \lambda/ a$$

    $$=\displaystyle \dfrac{500 \times 10^{-9}}{5 \times 10^{-3}} = 10^{-4} $$ radians

    Now, $$arc = angle \times radius = 10^{-4} \times (500\times 10^3)m = 50 m$$
  • Question 5
    1 / -0
    An observer moves with a velocity v towards a source. If actual velocity of light from source is c, its velocity relative to the observer is
    Solution
    Consider two particles with velocity u and w. The expression for the relative velocity can be given by

    $$v = \dfrac{w-u}{1-\dfrac{wu}{c^2}}$$.

    If one of the particle is light, $$w=c$$.

    This gives relative velocity of light $$v=c$$.

    Relative velocity of light in vacuum is always constant.

    Ans: B
  • Question 6
    1 / -0
    Wavelength of light used in an optical instrument are $$\lambda_1 = 4000 A^o  and \lambda_2 = 5000 A^0$$, then ratio of their respective resolving powers (corresponding to $$\lambda_1   \ and  \ \lambda_2$$) is
    Solution
    Resolving power of an optical instrument $$\displaystyle \propto \dfrac {1}{\lambda}$$

    $$\displaystyle \dfrac{\text{Resolving  power  at} \lambda_1} {\text{Resolving  power  at} \lambda_2}=\dfrac {\lambda_1}{\lambda_2}$$

    $$\displaystyle \left [\text{Limit of resolution}  \propto \dfrac{1}{\text{resolving  power}}\right]$$

    $$\therefore$$  Ratio  of  resolving  power $$= \displaystyle \dfrac {5000}{4000} = \dfrac {5}{4} = 5 : 4$$
  • Question 7
    1 / -0
    The colour of bright fringe nearest to central achromatic fringe in the interference pattern with white light will be 
    Solution
    As  $$\beta$$ = $$\displaystyle\frac{\lambda D}{d}$$ $$\therefore$$    $$\beta  \alpha   \lambda$$ 
    As $$\lambda$$ for violet is least, therefore, fringe nearest to central achromatic fringe will be violet   
  • Question 8
    1 / -0
    The interfering fringes formed by a thin oil film on water are seen in yellow light of sodium lamp. We find the fringes 
    Solution
    When the yellow light of sodium lamp interferes constructively we get yellow bright band, and when they interfere destructively we get black bark band.
  • Question 9
    1 / -0
    Light transmitted by nicol prism is
    Solution
    Nicol prism is a polariser in which the O-ray is eliminated by total internal reflection and the light transmitted through it, is E-ray which is completely plane polarised light.
  • Question 10
    1 / -0
    The waveforms of a light wave traveling in vacuum are given by $$x+y+z=c$$.The angle made by the direction of propagation of light with the $$X$$-axis is
    Solution
    Let any $$\vec{R}$$ its components are$$\vec{R}$$ = $$\vec{R}$$x + $$\vec{R}$$y + $$\vec{R}$$z 
    with | $$\vec{R}$$ | =$$\sqrt{Rx^{2} + Ry^{2} + Rz^{2} }$$

    & cos$$\theta_x = \displaystyle\frac{Rx}{R},  cos\theta_y =\frac{Ry}{R},  cos\theta_z = \frac{Rz}{R}$$

    there cos$$\theta_x$$, cos$$\theta_y$$ and cos$$\theta_z$$ one called direction cosines.

    Hence x + y + z = c $$(= \vec{R})$$
    So, magnitude of c = $$\sqrt{I^{2} + I^{2} + I^{2}} = \sqrt{3}$$ 
    and cos$$\theta_x$$ = $$\dfrac{1}{\sqrt{3}}$$
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