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Human Eye and Colourful World Test - 38

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Human Eye and Colourful World Test - 38
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  • Question 1
    1 / -0
    At sun rise or at sun set the sun appears to be reddish while at mid day it looks white. This is because
    Solution
    It is due to scattering of light. At noon, the sun is at the nearest point to earth. Hence all the colors get scattered almost equally and this you see white light. 
    But during sunrise and sunset, the sun is at the longest position. Here blue light gets scattered off and due to larger wavelength of the red light it can travel a longer distance and hence reach our eyes. This we see the sun red.

    Option A is correct.
  • Question 2
    1 / -0
    A person cannot see objects clearly which are nearer than 75 cms from his eyes, the disease he is suffering from is:
  • Question 3
    1 / -0
    A person can see clearly up to 3 metres what should be the power of the lens in his spectacles so that he could see clearly up to 12 metres.
    Solution
    Answer is B.

    The lens equation is given as $$\displaystyle \frac{1}{u}+ \frac{1}{v} = \frac{1}{f}$$
    $$\displaystyle \frac{1}{1200} + \frac{1}{-300} - \frac{1}{f}$$ (u and v are in cm) 

    The negative sign is due negative value of the image distance of the concave lens.
    $$f = - 400\ cm$$
    Power $$=\displaystyle \frac{100}{f}$$ in cm.
    So, $$\displaystyle \frac{100}{-400} = - 0.25 D$$.
    Hence, the power of the lens in his spectacles is -0.25D.
  • Question 4
    1 / -0
    Reflection of the sunlight from the upper layers of the Earth's atmosphere causes
    Solution
    You can define twilight simply as the time of day between daylight and darkness, whether that’s after sunset, or before sunrise. It’s a time when the light from the sky appears diffused and often pinkish. The sun is below the horizon, but its rays are scattered by Earth’s atmosphere to create the colors of twilight.

    We have twilight because Earth has an atmosphere. Some light scatters through small particles in the atmosphere – so there’s still some light in the sky even after the sun has gone down.
  • Question 5
    1 / -0
    Figure (a), (b), and (c) respectively, indicate the point of focus in case of :

    Solution
    Normal eye forms an image at retina. A myopic eye forms an image before retina whereas a hypermetropic eye forms image behind retina.
  • Question 6
    1 / -0
    Long-sightedness is caused by the eyeball being too short. It can be corrected by the use of a :
    Solution
    In long sightedness, size of the eyes ball decrease due to which image of the object formed behind the ratina. This defect can be corrected by using additional converging lens of sufficient power. That lens will help the eyes to formed image on the ratina. Hence converging lens required. 
  • Question 7
    1 / -0
    Hypermetropia is due to the ________ of the eye.
    Solution
    Hypermetropia, sometimes called hyperopia, is the term used to define being longsighted. In hypermetropia, the image of a nearby object form behind the retina. This means that light is focused too far back in the eye. This happen due to decrement in converging power of eyes lens. 
    Hence, Hypermetropia is due to the low converging power of the eye.
  • Question 8
    1 / -0
    A person cannot see the objects clearly placed at distance more than $$40\ cm$$. He is advised to use lens of power:
    Solution
    That lens should be given to the person which forms the image of an object at infinity at distance 40 cm in front of the eye.

    Hence, $$u=-\infty$$ and $$v=-40cm$$

    Now,

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

    $$\implies \dfrac{1}{f}=\dfrac{-1}{40}-\dfrac{1}{-\infty}$$

    $$\implies f=-40cm$$

    $$P=\dfrac{100}{f(cm)}=\dfrac{-100}{40}=$$

    $$\implies P=-2.5D$$

    Answer-(A).
  • Question 9
    1 / -0
    A near sighted person cannot see distinctly beyond $$50   cm$$ from his eye. The power in diopter of spectacle lenses which will enable him to see distant objects clearly is:
    Solution

    When a person cannot see far situated object we assume object distance to infinity.

    Then image distance be v = -50cm

    $$u=\infty \\ v=-50cm$$

    then,

    $$\dfrac { 1 }{ f } =\dfrac { 1 }{ v } -\dfrac { 1 }{ u } \\ \dfrac { 1 }{ f } =-\dfrac { 1 }{ 50 } \\ f=-50cm=-0.5m$$    (concave lens)

    $$power=\dfrac { 1 }{ f } =\dfrac { 1 }{ 0.5 } =-2D$$

  • Question 10
    1 / -0
    Figures (a), (b), (c) and (d) respectively correspond to 

    Solution
    In Myopia eyes, Images of an object form before the retina so for correction diverging lens are used.
     In Hypermetropia eyes, Images of an object formed behind the retina so for correction converging lens are used.
    Hence,
    R.E.F image
    (a) Myopia of short-sightedness 
    R.E.F image
    (b) Corrected short-sightedness
    R.E.F image
    (c) Hypermetropia or long-sightedness 
    R.E.F image
    (d) Corrected hypermetropia 

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