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Human Eye and Colourful World Test - 50

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Human Eye and Colourful World Test - 50
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Weekly Quiz Competition
  • Question 1
    1 / -0
    Which colour of light is scattered the most (maximum) in the atmosphere?
    Solution
    The molecules of air and other fine particles in the atmosphere have their size smaller than the wavelength of visible light. Thus, they are more effective in scattering light of shorter wavelengths, e.g. blue light. The red light has a larger wavelength than blue light. Thus, when sunlight is passing through the atmosphere, the fine particles in air scatter the blue colour (shorter wavelengths) more strongly than red. 
  • Question 2
    1 / -0
    The lens used to correct long sightedness or hypermetropla.
    Solution
    A person suffering from hypermetropia can see only distant objects clearly but can't see the nearby objects. Hypermetropia can be corrected by using a convex lens.
  • Question 3
    1 / -0
    How many times do the components of white light undergo deviation while passing through the prism ?
    Solution
    When a ray of light enters the glass prism it gets deviated two times. First when it enters the glass prism and second when it comes out of the prism. This is because the refracting surfaces of the prism are not parallel to each other. Also, when the ray of light passes through the prism it bends towards its base.

  • Question 4
    1 / -0
    In spectrum obtained with prism which color is deviated maximum ?
    Solution
    Violet colour having higher refractive index and hence undergoes maximum deviation.

    Answer-(A)
  • Question 5
    1 / -0
    A person near point of vision is $$100 cm$$. Then, the power of lens he must wear so as to have normal vision is :
    Solution
    $$ f = \dfrac {yD} {y-D} = \dfrac {100 \times 25 }{100 - 25 } = \dfrac {100}{3} cm = \dfrac  {1}{3} m $$
    $$ P = \dfrac  {1}{f} = + 3 D $$
  • Question 6
    1 / -0
    Dispersion of light is caused due to :
  • Question 7
    1 / -0
    In the dispersion of white light from a prism, the violet light is deviated more than the red light because :
    Solution
    In the dispersion of white light from a prism, the violet light is deviated more than the red light because the refractive index of prism for violet rays of light is more.
  • Question 8
    1 / -0
    For the myopic eye, the defect is cured by
    Solution
    Myopia is a vision condition, in which a person can see nearby objects clearly but objects farther away appear blurry. Hence the far point of the eye is reduced.

    So to cure this , we should use a lens of focal length $$f$$ such that it forms the image of objects at infinity at the distance of the far point of the person.

    Hence, applying the lens' formula

    In myopia, $$\displaystyle u=\infty ,v= -d=$$ distance of far point
    By $$\displaystyle \frac { 1 }{ F } =\frac { 1 }{ v } -\frac { 1 }{ u } $$, we get $$\displaystyle f=-d$$
    Since f is negative, hence the lens used is concave.

  • Question 9
    1 / -0
    A person can see objects clearly only upto a maximum distance of $$60\ cm$$. His eye defect, nature of the corrective lens and its focal length are respectively.
    Solution
    Range of vision for healthy eye is $$25\ cm$$ to $$\infty$$. If the person can see clearly only upto a maximum distance of $$60\ cm$$. He is suffering from myopia. A short sighted eye can see only nearer objects. This defect can be removed by using a concave lens of suitable focal length equal to maximum distance seen by person. $$f = 60\ cm$$.
  • Question 10
    1 / -0
    A far-sighted person has his near point $$50\ cm$$, find the power of lens he should use to see at $$25\ cm$$, clearly.
    Solution
    Given: A farsighted person has his near point 50cm, 
    To find the power of lens he should use to see at 25cm, clearly.
    Solution:
    Distant objects need to be imaged at most 50 cm from the eye.
    According to the given criteria,
    $$u=25cm, v=-50cm$$
    Applying lens formula, we get
    $$\dfrac 1f=\dfrac 1v+\dfrac 1u\\\implies \dfrac 1f=\dfrac 1{-50}+\dfrac 1{25}\\\implies \dfrac 1f=\dfrac {-1+2}{50}\\\implies f=50cm=0.5m$$
    The power of the lens he should use is,
    $$P=\dfrac 1f=\dfrac 1{0.5}=+2D$$
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