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Human Eye and Colourful World Test - 26

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Human Eye and Colourful World Test - 26
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Weekly Quiz Competition
  • Question 1
    1 / -0
    Which of the following statements is correct?
    Solution
    Myopia is also termed as short-sightedness. A person suffering from myopia can see nearby objects clearly but not distant objects.
    Whereas hypermetropia is termed as long-sightedness. A person suffering from hypermetropia can see distant objects clearly but not the nearby objects.  
    Hence Option A
  • Question 2
    1 / -0
    What is the least distance of distinct vision of a normal human eye?
    Solution
    The least distance of distinct vision of a normal eye is 25 cm.
  • Question 3
    1 / -0
    If there had been one eye of the man, then 
    Solution
    The pair of eyes are responsible for an increase in the field of vision and also adding depth to our vision, i.e. images becomes three dimensional.
    Hence, option D is correct.
  • Question 4
    1 / -0
    If the distance of the far point for a myopia patient is doubled, the focal length of the lens required to cure it will become 
    Solution
    Focal length = $$-$$( Detected far point )
    For curing myopia, a concave lens of focal length equal to the distance of the eye’s own far point is used.  i.e. $$f = − d$$  Hence, if distance d is doubled, then the focal length will also be doubled.
  • Question 5
    1 / -0
    The sensation of vision in the retina is carries to the brain by 
    Solution
    The sensation of vision in the retina is carried to the brain by the Optic nerve. 
    Once light reflected from the surrounding object enters into retina through eye lens, chemical and electrical impulses (sensory information) are generated. and the optic nerve transmits visual information to the brain via electric impulses.
  • Question 6
    1 / -0
    The far point of the myopic eye is $$250\ cm$$, then the focal length of the lens to be used will be 
    Solution
    The correcting lens of focal length $$250\ cm$$, since myopic eye is corrected by using a concave lens (which is a diverging lens) of suitable power.
  • Question 7
    1 / -0
    When the power of eye lens increases, the defect of vision is produced. This  defect of vision is known as 
    Solution
    When the power of eye lens increases, the object kept at the far point can not be seen clearly because the focal distance of eye lens decreases. This type of defect of vision is known as Shortsightedness.
  • Question 8
    1 / -0
    A student can distinctly see the object upto a distance $$15\ cm$$. He wants to see the black board at a distance of $$3\ m$$. Focal length and power of lens used respectively will be 
    Solution
    $$v=-15\ cm, u=-300\ cm$$
    From lens formula $$\dfrac 1f=\dfrac 1v-\dfrac 1u$$
    $$\Rightarrow \dfrac 1f=\dfrac{1}{-15}-\dfrac {1}{-300}=\dfrac{-19}{300}\Rightarrow f=\dfrac {-300}{19}=-15.8\ cm$$
    and power $$P=\dfrac {100}{f}\ cm=\dfrac{-100 \times 19}{300}=-6.33\ D$$.
  • Question 9
    1 / -0
    The power of a lens used to remove the myopic defect of an eye is 0.66 D. The far point for this eye is (nearly) :
    Solution
    As $$p= \dfrac{1}{f}$$

    So, $$f= \dfrac{1}{p}$$

    $$\Longrightarrow$$  $$f= \dfrac{100}{0.66}cm$$

    $$\Longrightarrow$$ $$f= 151.51cm$$

    $$\Longrightarrow$$ $$f\simeq 150cm$$
  • Question 10
    1 / -0
    A person can see clearly objects between 15 and 100 cm from his eye. The range of his vision if he wears close fitting spectacles having a power of 0.8 diopter is :
    Solution
    Given, $$v= \infty $$

    Focal length, $$f= \dfrac{1}{p}$$

    $$= \dfrac{100}{0.8}$$

    $$= 125 cm$$

    To see the objects before 15 cm 

    $$v= +15\ ,\ u=?$$

    $$\dfrac{1}{v}-\dfrac{1}{u}= \dfrac{1}{f}$$

    $$\dfrac{1}{15}-\dfrac{1}{u}= \dfrac{1}{125}$$

    $$u= 17.04\ cm$$

    To see the object far away from 100 cm

    $$u= 100cm,\ v= ?$$

    $$\dfrac{1}{v}-\dfrac{1}{u}= \dfrac{1}{f}$$

    $$\dfrac{1}{v}-\dfrac{1}{100}= \dfrac{1}{125}$$

    $$v= \ 500cm$$
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