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Human Eye and Colourful World Test - 56

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Human Eye and Colourful World Test - 56
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  • Question 1
    1 / -0
    The eye defect represented by the figure is

    Solution
    In the given figure, Diverging lens are used for clear viewing. Its means that defected eyes was making image of an object inform of ratina. This defect of eyes is known as Myopia.
    Option A
  • Question 2
    1 / -0
    The color of the sky is blue during the daytime, red during sunset, and black at night. This is due to
    Solution
    A and C. Gases and particles in Earth's atmosphere scatter sunlight in all directions. Blue light is scattered more than other colors because it travels as shorter, smaller waves. This is why we see a blue sky most of the time.
    B. During sunrise and sunset the sun is low in the sky, and it transmits light through the thickest part of the atmosphere. We see the red because red wavelengths (the longest in the color spectrum) are breaking through the atmosphere. The shorter wavelengths, such as blue, are scattered and broken up.

  • Question 3
    1 / -0
    A student of class $$ 10, $$ is not able to see clearly the blackboard question when seated at a distance of $$ 5\  \mathrm{m} $$ from the board, the defect he is suffering from is
    Solution
    Myopia is a vision condition in which you can see nearby objects clearly but far away objects appear blurry. In this defect, the light from far away objects focusses in front of the retina. 

    Since, the student is not able to see blackboard, which is located at a far away($$5 \ m$$) distance, clearly. Hence, he is suffering from Myopia.
  • Question 4
    1 / -0
    Long-sightedness or hypermetropia can be corrected by
    Solution
    Convex lens
  • Question 5
    1 / -0
    The colored light that refracts most while passing through a prism is
    Solution
    The white light is dispersed into its seven-color components by a prism.  Different colors of light bend through different angles with respect to the incident ray, as they pass through a prism. The red light bends the least while the violet the most.
  • Question 6
    1 / -0
    A man can see the object up to a distance of one meter from his eyes. For correcting his eyesight so that he can see an object at infinity, he requires a lens whose power is 
    Solution
    The image of object at infinity should be formed at $$100\ cm$$ from the eye 
    Apply formula:
    $$\dfrac 1f =\dfrac{1}{v}-\dfrac{1}{u}$$
    $$f$$ : focal length
    $$u$$ : object distance
    $$v$$ : Image distance
    $$\dfrac 1f =\dfrac{1}{\infty}-\dfrac{1}{100}$$
    $$\Rightarrow \dfrac{1}{f}=-\dfrac{1}{100}$$
    Apply formula of power:
    $$P = \dfrac{1}{f}$$
    So the power $$=\dfrac{-100}{100}=-1\ D$$
    ( Distance is given in $$cm$$ but $$P=\dfrac 1f$$ in metres )
  • Question 7
    1 / -0
    A man can see the object between $$15\ cm$$ and $$30\ cm$$. He uses the lens to see the far objects. Then due to the lens used, the near point will be at 
    Solution
    For improving far point, concave lens is required and for this concave lens $$u=\infty, v=-30\ cm$$
    So,
     $$\dfrac 1f=\dfrac{1}{-30}-\dfrac{1}{\infty}$$
    $$\Rightarrow f=-30\ cm$$
    for near point,
    $$\dfrac{1}{-30}=\dfrac{1}{-15}-\dfrac 1u $$
    $$\Rightarrow u=-30\ cm$$
  • Question 8
    1 / -0
    A man suffering from myopia can read a book placed at $$10\ cm$$ distance. For reading the book at a distance of $$60\ cm$$ with relaxed vision, the focal length of the lens required will be:
    Solution
    According to the question,
    Far point of the myopic eye $$= 10\ cm$$
    Far point of a normal eye $$= 60\ cm$$
    object distance, $$u = -60\ cm$$
    image distance, $$v = -10\ cm$$
    focal length, $$f$$
    Using lens formula
    $$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$$

    $$\dfrac{1}{f} = \dfrac{1}{-10} - \dfrac{1}{\left(-60\right)} = \dfrac{-6 + 1}{60}$$

    $$\dfrac{1}{f} = \dfrac{-5}{60}$$

    $$f = -12\ cm$$
  • Question 9
    1 / -0
    The far point of a myopia eye is at $$40\ cm$$. For removing this defect, the power of lens required will be 
    Solution
    For myopic eye $$f=-$$ ( defect far point )
    $$\Rightarrow f=-40\ cm\Rightarrow P=\dfrac{100}{-40}=-2.5\ D$$
  • Question 10
    1 / -0
    When white light passes through the achromatic combination of prisms, then what is observed
    Solution
    When white light passes through a prism , it disperses into band of seven colours. But when two prism are combined in such a way that sum of angular dispersions of crown glass prism and flint glass prism is zero then such a combination is achromatic combination of prisms. When white light passes through achromatic combination of prisms, internally dispersed components of white light from Crown glass prism refract and meet together at the outer surface edge of Flint Glass prism ,so the refracted light from the achromatic combination of prisms become white light again with deviation only without any dispersion. 

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