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Human Eye and Colourful World Test - 19

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Human Eye and Colourful World Test - 19
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Weekly Quiz Competition
  • Question 1
    1 / -0
    The power of lens in the spectacles of a person is $$+2D$$. The person suffers from :
    Solution
    As power is $$+2D$$, therefore focal length is $$= \dfrac{1}{P}$$
    which is $$+\dfrac{1}{2}m=+50cm$$ so, focal length is positive which is for convex lens and a convex lens is used for correction of hypermetropia. So, the person is suffering from hypermetropia, hence correct option is $$A$$.
  • Question 2
    1 / -0
    The focal length of a normal eye-lens is about:
    Solution
    The focal length of a normal eye lens is about $$2\ cm$$ because distance between eye lens and retina is approximately $$2\ cm$$. So, when parallel rays are coming they form image at focus and to make image at retina focal length should be about $$2\ cm$$.
  • Question 3
    1 / -0
    Inability of the eye to focus on both far and near objects with advancing age is:
    Solution
    With advancing age when ciliary muscles become weak and are not able to strain enough to reduce focal length to appropriate value then this defect is known as presbyopia.
  • Question 4
    1 / -0
    In the case of hypermetropia:
    Solution
    In hypermetropia image of a distance object is formed on the retina and image of a near object is formed behind the retina. For correction of hypermetropia , convex lens is used.
  • Question 5
    1 / -0
    A myopic person cannot see distinctly , objects that are :
    Solution
    Myopia is a vision defect in which a person can see nearby objects clearly but is not able to see distinct objects clearly. In this defect, the rays coming from distant objects, focus in front of retina, instead of on retina.

  • Question 6
    1 / -0
    When objects at different distances are seen by the eye, which of the following remains constant?
    Solution
    In the eye, image is always formed at retina which is at a fixed distance in eye. So, when objects at different distances are seen by the eye, the image distance from the eye lens remains constant by changing radii of curative of the eye lens leading to change in focal length. As objects are at different distances, object distance from eye lens changes.
  • Question 7
    1 / -0
    What happens when white light passes through a glass prism?
    Solution
    Upon passage through the prism, the white light is separated into its component colors - red, orange, yellow, green, blue, indigo and violet.
  • Question 8
    1 / -0
    The power of lens, a short sighted person uses is $$- 2 \ D$$. The maximum distance of an object which he can see without spectacles is 
    Solution
    The maximum distance of an object, which a person can see without spectacles is equal to the focal length of the lens.

    Hence, let $$ focal \ length = f$$
                       $$Power = P = -2 \ D$$

    $$f= \dfrac{100}{P}$$, 

    $$= \dfrac{100}{-2}cm$$

    $$= -50\ cm$$

    So he can see upto a distance of $$50 \ cm$$, without spectacles. 
  • Question 9
    1 / -0
    A myopic person can not see objects lying beyond 2m. The focal length and power of the lens required to remove this defect will be?
    Solution
    $$u=+2m$$

    $$v= \infty $$

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

    $$\dfrac{1}{\infty }-\dfrac{1}{2}=\dfrac{1}{f}$$

    $$f= -2m$$

    $$P= \dfrac{1}{f}$$

    $$= \dfrac{1}{-2}$$

    $$= -0.5 D$$
  • Question 10
    1 / -0
    A long sighted person has a least distance of distinct vision of 50 cm. He wants to reduce it to 25 cm. He should use a :
    Solution
    Given, $$u= \infty $$

    Image distance, $$v= 50cm$$

    From lens formula, $$\dfrac{1}{v}-\dfrac{1}{u}= \dfrac{1}{f}$$

    $$\dfrac{1}{50}-\dfrac{1}{\infty }= \dfrac{1}{f}$$

    $$f= +50\ cm$$

    He should use a convex lens of focal length 50 cm.
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