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Human Eye and Colourful World Test - 52

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Human Eye and Colourful World Test - 52
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  • Question 1
    1 / -0
    The nearer point of hypermetropic eye is 40 cm. The lens to be used for its correction should have the power?
    Solution
    Near point of a hypermetropic eye =40 cm = 40\ cm  
    Near point of a normal eye =25 cm = 25\ cm
    Object distance, u=25 cmu = -25\ cm
    Image distance, v=40 cmv = -40\ cm
    Using lens formula,
    1f=1v1u=140+125 \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{-40} + \dfrac{1}{25}

    1f=5+8200=3200 cm=300200 m\dfrac{1}{f} = \dfrac{-5 + 8}{200} = \dfrac{3}{200\ cm} = \dfrac{300}{200\ m}

    P=1f(in m)=300200=32P = \dfrac{1}{f {\text{(in m)}}} = \dfrac{300}{200} = \dfrac{3}{2}

    P=1.5 DP = 1.5\ D
  • Question 2
    1 / -0
    In the given figure, if P is a point of the source of light, R is retina and L is the lens of the eye, the person having such condition of his eyes is suffering from :

    Solution
    In the given figure, Image of a point P is forming behind the retina. this defect of eyes is known as Hypermetropia. Hypermetropia means farsightedness. In this defect, the focal length increases or size of eyeball decreases and image is formed beyond the retina. This defect is corrected by using a convex lens.

  • Question 3
    1 / -0
    Which of the following statements is correct concerning the passage of white light into a glass prism?
    Solution
    Violet colour shows more deviation as slower light shows more deviation. Hence, violet is slower than red. Red deviates less and violet deviates more.
  • Question 4
    1 / -0
    Of the following natural phenomena, tell which one is based on the refraction of light:
    Solution

  • Question 5
    1 / -0
    A ray of light containing both red and blue colours is incident on the refracting surface of a prism. Then,
    Solution
    Different colours of light bend through different angles with respect to the incident ray, as they pass through a prism. The light ray with more wavelength suffers less deviation than light ray with more wavelength. So the red light bends the least while the blue the most.

    wavelength of red > wavelength of blue.

    So option (c): "red colour suffers less deviation" is correct.
  • Question 6
    1 / -0
    A person cannot see object clearly that are closed that 2m2m and father than 4m4m . To correct the eye vision the person will use :
    Solution
    Given,
    The distance of the far point (D)=2m(D) = 2m
    Let f be the focal length of the eye lens.
    Image distance ( distance between eye lens and retina ) for our human eye is always constant and is about 
     0.25m.0.25m.
    This is due to the action of ciliary muscles. 
    applying lens formula, we get 
    f=D=2m.f = -D = -2 m.
    But, the power of a lens is reciprocal of focal length.
    Hence. Power =1f.=\dfrac{1}{f}. ( f in meters )
    12=0.5\dfrac{1}{-2}=0.5D
    Similarly,
    The distance of the far point (D)=4m(D) = 4m
    Let f be the focal length of the eye lens.
    Image distance ( distance between eye lens and retina ) for our human eye is always constant and is about 
     0.25m.0.25m.
    This is due to the action of ciliary muscles.
    applying lens formula, we get 
    f=D=4m.f = -D = -4 m.
    But, the power of a lens is reciprocal of focal length.
    Hence. Power =1f.=\dfrac{1}{f}. ( f in meters )
    14=0.25\dfrac{1}{-4}=0.25D
    So he has to use by focal lenses
  • Question 7
    1 / -0
    A man wearing glasses of focal length + 1m cannot clearly see beyond 1 m
    Solution

  • Question 8
    1 / -0
    To remove myopia (short-sightedness) a lens of power 0.66D- 0.66D is required. The distant point of the eye is approximate.
    Solution
    The given power(P)(P) is 0.66D-0.66D.
    We know that-
    P=1fP = \dfrac{ 1 }{ f }
    ff; focal length 

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

    f=10.66\Rightarrow f = \dfrac{ 1 }{ -0.66 }

    f=151.5 cm\Rightarrow f = 151.5\ cm

    Hence, the distant point of the eye is 151.5 cm151.5\ cm.
  • Question 9
    1 / -0
    A man wearing glasses of power +2D+2D, can read clearly a book placed at a distance of 40 cm40\ cm, from the eye. The power of the lens required, so that he can read at 25 cm25\ cm from the eye is ?

    Solution
    P=2P = 2
    f=1P=12 m=50 cm\Rightarrow f = \dfrac{1}{P} = \dfrac{1}{2}\ m = 50\ cm
    u=40cmu=-40cm

    1f=1v1u\Rightarrow \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}  

    150=1v140\Rightarrow \dfrac{1}{50} = \dfrac{1}{v} - \dfrac{1}{-40}  

    150140=1v\Rightarrow \dfrac{1}{50} - \dfrac{1}{40} = \dfrac{1}{v}  

    v=200 cm\Rightarrow v = -200\ cm

    Now for new lens object distance = new near point =-25cm25cm
    u2=25 cmu_{2} = -25\ cm
    v2=200 cmv_{2} = -200\ cm

    1f=1v21u2\Rightarrow \dfrac{1}{f} = \dfrac{1}{v_2} - \dfrac{1}{u_2}  

    =1200125 = \dfrac{1}{-200} - \dfrac{1}{-25}  

    1f=1251200=7200 cm1=72 m1 \dfrac{1}{f} = \dfrac{1}{25} - \dfrac{1}{200} = \dfrac{7}{200}\ cm^{-1} = \dfrac{7}{2}\ m^{-1} 

    P=1fP =\frac{1}{f}  = +3.5 D+3.5\ D
  • Question 10
    1 / -0
    A far-sighted person cannot focus distinctly objects closer than 120 cm. The lens that will permit him to read from a distance of 40 cm will have a focal length:
    Solution

    Given,

    Near point, v=120cmv=-120\,cm

    Reading Distance, u=40cmu=-40\,cm

    From lens formula,

      1f=1v1u=1120140=160 \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{-120}-\dfrac{1}{-40}=\dfrac{1}{60}

     f=+60cm f=+60\,cm

    Focal length of lens f=+60cmf=+60\,cm 

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