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Human Eye and Colourful World Test - 48

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Human Eye and Colourful World Test - 48
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  • Question 1
    1 / -0
    Due to the different frequency of each colour, all components of white light get _____ at a different angle.
    Solution
    The index of refraction of a medium depends upon the frequency of the wave refracted. Since, different colours have different frequencies and wavelengths.

    Hence,  the components of the white light refract at different angles depending upon their frequencies.

  • Question 2
    1 / -0
    Which of the following observations cannot be explained by Tyndall Effect?
    Solution
    Tyndall Effect is scattering of light in a medium containing small suspended particles. It is observed in many natural phenomena. Some common phenomena include the sunlight entering through forest canopy, sunlight entering a dark room from a small hole and light passed through a milk solution. 

    Since, the Tyndall effect is observed in a medium containing fine suspended particles. It cannot explain that the speed of light in vacuum is $$3 \times 10^8\ m/s$$.

    Hence, the correct option is C.

  • Question 3
    1 / -0
    Tyndall effect is shown by:
    Solution
    The Tyndall Effect (Tyndall scattering), is the scattering of light by particles in a colloid or the particles in a very fine suspension. It is observed in aerosol, milk, etc.
  • Question 4
    1 / -0
    Which of the following phenomenon is involved in Tyndall effect?
    Solution
    The Tyndall Effect (Tyndall scattering), is light scattering by particles in a colloid or else particles in a very fine suspension. The amount of scattering is inversely proportional to the fourth power of wavelength and hence the colour of the colloidal solution is the same as the least coloured wavelength.
  • Question 5
    1 / -0
    Purple light has a higher index of refraction than green light which has a higher index of refraction than red light.
    If the black ray represents white light traveling into the left side of the pictured prisms, which prism most accurately shows the colors of light as they come out of the prism?
    Solution
    When light passes through the prism, the particles of purple colors( shorter wavelength) more strongly scattered or deviated  than green light which is more strongly scattered than the red light (higher wavelength). Hence the answer is option  C.
  • Question 6
    1 / -0
    In the following ray diagram the correctly marked angles are.

    Solution
    $$\angle A$$ represents the angle of a prism while $$\angle D$$ represents the angle of deviation and both are  correctly marked.
    As angle of refraction $$(\angle r)$$ is the angle formed by the refracted ray with the normal and is also correctly marked.
    But $$\angle i$$ and $$\angle e$$ are the angles made by the incident ray with the normal and emergent ray with the  normal respectively and thus are incorrectly marked.
    Hence option D is correct.
  • Question 7
    1 / -0
    If the image of distant objects is formed in front of the retina, the defect of vision may be
    Solution
    In nearsightedness, also called myopia, the image of a distant object is formed in front of the retina and not at the retina. This defect arises because the power of the eye is too large due to the decrease in focal length of the crystalline lens.
  • Question 8
    1 / -0
    Red light is used as danger signal because it is:
  • Question 9
    1 / -0
    The inability among the elderly to see nearby objects clearly because of the weaking of the ciliary muscles is called
    Solution
    Presbyopia
    This is caused by loss of elasticity of the lens of the eye, occurring typically in middle and old age.The ciliary muscles become weak and can not adjust shape of lens any more to change focal length of eye lens so the person can not see nearby objects clearly.
  • Question 10
    1 / -0
    Far point of a person is $$5 m$$. In order that he has normal vision what kind of spectacles should be use
    Solution
    $$x-$$ distance of far point $$=5  meters$$
    $$f=$$focal length of correcting lens
    For lens, $$v=$$image distance$$=-x$$ ,$$u=$$object distance $$=\alpha$$
    $$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$$
    $$=\dfrac{1}{-x}+\dfrac{1}{\alpha}$$
    So $$f=-5m$$
    Since it is negative ,it is concave lens.
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