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Human Eye and Colourful World Test - 57

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Human Eye and Colourful World Test - 57
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  • Question 1
    1 / -0
    A person is suffering from myopic defect. He is able to see clear objects placed at 15 cm15\ cm. What type and of what focal length of lens he should use to see clearly the objects placed 60 cm60\ cm away 
    Solution
    Lens should be used such that the image of an object placed at 60 cm60 \ cm is formed at 15 cm15 \ cm

    Hence, in the given case u=60 cmu=-60 \ cm; v=15 cmv=-15 \ cm 
    Using the lens' formula 1f=1v1u\dfrac 1f=\dfrac{1}{v}-\dfrac{1}{u}

    1f=115160=360\Rightarrow \dfrac 1f=\dfrac{1}{-15}-\dfrac{1}{-60} = \dfrac{-3}{60}

    f=20 cm\Rightarrow f=-20\ cm.

    Hence, the person should use concave lens of focal length 20 cm20 \ cm

  • Question 2
    1 / -0
    The hyper-metropia is a 
    Solution
    The hyper-metropia is an eye defect when person faces difficulty in seeing nearby objects(Short Sight) but not far-sided objects (Long Sight). Due to inability of the eye lens to converge light rays from nearby objects at the retina, image is formed behind the retina. Long sight leads to problems with near vision. So hyper-metropia is a Long-side defect.
  • Question 3
    1 / -0
    At sun rise of sunset, the sun looks more red than at mid-day because
    Solution
    I1λ4I\propto\dfrac{1}{{\lambda}^4}
    According to Rayleigh’s law of scattering, intensity scattered is inversely proportional to the forth power of wavelength. So red is least scattered and sun appears Red.
  • Question 4
    1 / -0
    Stars are twinkling due to
    Solution
    Stars twinkle due to variation in R.lR.l of atmosphere.
    Stars are twinkling due to refraction,
  • Question 5
    1 / -0
    A person is suffering from 'presbyopia' ( myopia and hyper metropia both defects ) should use 
    Solution
    A bifocal lens consists of both convex lenses with lower part is convex.
  • Question 6
    1 / -0
    A person uses a lens of power +3 D+3\ D to normalise vision. Near point of hypermetropic eye is 
    Solution
    Focal length of the lens f=1003 cmf=\dfrac{100}{3}\ cm
    By lens formula 1f=1v1u\dfrac 1f=\dfrac 1v-\dfrac 1u
    1+100/3=1v125v=100 cm=1 m\Rightarrow \dfrac{1}{+100/3}=\dfrac 1v-\dfrac {1}{-25}\Rightarrow v=-100\ cm=-1\ m
  • Question 7
    1 / -0
    A person wears glasses of power 2.5 D-2.5\ D. The defect of the eye and the far point of the person without the glasses are respectively 
    Solution
    Negative power is given, so defect of eye is nearsightedness 
    Also defected far point =f=1p=100(2.5)=40 cm=-f=-\dfrac 1p=-\dfrac{100}{(-2.5)}=40\ cm
  • Question 8
    1 / -0
    To remove myopia ( short sightedness ) a lens of power 0.66 D0.66\ D is required. The distant point of the eye is approximately 
    Solution
    Given,
    Power of lens P=0.66 DP=0.66\ D
    Far point of the eye == focal length of the lens =100P=1000.66=151 cm=\dfrac{100}{P}=\dfrac{100}{0.66}=151\ cm
    Hence, The distant point of the eye is approximately 151 cm151\ cm
  • Question 9
    1 / -0
    Colour of the sky is blue due to
    Solution
    The molecules of air and other fine particles in the atmosphere have size smaller than the wavelength of visible light. These are more effective in scattering light of shorter wavelengths at the blue end than light of longer wavelengths at the red end. Thus, when sunlight passes through the atmosphere, the fine particles in air scatter the blue colour (shorter wavelengths) more strongly than red. The scattered blue light enters our eyes.
  • Question 10
    1 / -0
    The band of seven colours that constitute white light is known as
    Solution
    A prism separates white light into a group of seven colors called a spectrum. These seven colors are always in the same order. The colors of the spectrum are red, orange, yellow, green, blue, indigo, and violet.

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