Tag: human eye and colourful world

Questions Related to human eye and colourful world

What does the dispersion of light in a medium imply
i) Lights of different wavelength travel with different speeds in the medium.
ii) Light of any frequency will travel with the same speed in the medium
iii) The refractive index of the medium is different for different wavelength of light

physics human eye and colourful world sunlight - white and coloured sunlight - combination of 7 lights fibre optics basics

  1. $(i)$ and $(ii)$ only

  2. $(ii)$ and $(iii)$ only

  3. $(i)$ and $(iii)$ only

  4. $(i),(ii)$ and $(iii)$

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Correct Option: C

A person cannot see object clearly that are closed that $2m$ and father than $4m$ . To correct the eye vision the person will use :

structure of human eye human eye and colourful world

  1. Bifocal lenses of power $0.5D$ and $0.25D$

  2. Bifocal lenses of power $0.25 D$ and $3.5D$

  3. Bifocal lenses of power $0.5 D$ and $4.0 D$

  4. Bifocal lenses of power $4.0 D$ and $0.5 D$

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Correct Option: A
Explanation:

Given,

The distance of the far point $(D) = 2m$
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 
 $0.25m.$
This is due to the action of ciliary muscles. 
applying lens formula, we get 
$f = -D = -2 m.$
But, the power of a lens is reciprocal of focal length.
Hence. Power $=\dfrac{1}{f}.$ ( f in meters )
$\dfrac{1}{-2}=0.5$D
Similarly,
The distance of the far point $(D) = 4m$
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 
 $0.25m.$
This is due to the action of ciliary muscles.
applying lens formula, we get 
$f = -D = -4 m.$
But, the power of a lens is reciprocal of focal length.
Hence. Power $=\dfrac{1}{f}.$ ( f in meters )
$\dfrac{1}{-4}=0.25$D
So he has to use by focal lenses

To remove myopia (short sightedness) a lens of power $- 0.66D$ is required. The distant point of the eye is approximately

structure of human eye human eye and colourful world

  1. $100$ cm

  2. $151.5$ cm

  3. $50$ cm

  4. $25$ cm

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Correct Option: C

Minimum and maximum distance should be for clear vision of healthy eye.

structure of human eye human eye and colourful world

  1. 100 cm & 500 cm

  2. Infinite & 25 cm

  3. 25 cm & 100 cm

  4. 25 cm & infinite

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Correct Option: D
Explanation:

Minimum distance for clear vision of healthy eye is  $25 \ cm$ and maximum distance is infinite.

Diameter of a human eye lens is 2 millimetre. What will be the minimum distance between two points to resolve, which are situated at the distance of 50 m from the eye? The wavelength of light is 5000 angstrom.

structure of human eye human eye and colourful world

  1. 2.32 m

  2. 4.28 mm

  3. 1.25 cm

  4. 12.48 cm

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Correct Option: B

State whether true or false:

A concave lens is used to correct short-sightedness in eye.

structure of human eye human eye and colourful world

  1. True

  2. False

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Correct Option: A
Explanation:

Answer is A.

A person who is short sighted can focus clearly on near objects but cannot focus on distant objects. This is because the eyeball is too long. Light from distant objects is focused at a point in front of the retina resulting in a blurred image.
This defect can be corrected by wearing a concave (diverging) spectacle lens. The rays of light from a near object are diverged before entering the eye so that the cornea and eye lens can direct the focal point onto the retina.
Hence, the statement is true.

A person cannot see the object beyond 200cm. The power of lens  correct the vision will be :

structure of human eye human eye and colourful world

  1. +0.5 D

  2. +5D

  3. -0.5D

  4. -5D

Show answer
Correct Option: C
Explanation:

Here, $u = \infty$ and $ v = -200\; cm$


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

$\Rightarrow \dfrac{1}{f} =- \dfrac{1}{200} + 0$

$\Rightarrow f =-2m$
$\Rightarrow  P (in \; D)= \dfrac{1}{f(in\; m)} =- 0.5 D$

Therefore, C is correct option.

The near point and the far point of a child are at 10 and 100 cm. If the retina is 2.0 cm behind the eyelens  .What is the range of the power of the eye lens?

structure of human eye human eye and colourful world

  1. $50 D \text { to } 40 \mathrm { D }$

  2. $60 D \text { to } 51 \mathrm { D }$

  3. $60 D \text { to } 54 \mathrm { D }$

  4. $40 D \text { to } 50 \mathrm { D }$

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Correct Option: A
Explanation:

\begin{array}{l} According\, \, to\, question................... \ Near\, point\, of\, the\, child\, (u)=10cm \ Far\, point\, of\, the\, child\, (u)=100cm \ and,the\, \, retina\, is\, \, \, 2\, cm\, behind\, the\, eye\, lens. \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, V=2\, cm=0.02\, m \ Now, \ \frac { 1 }{ f } =\frac { 1 }{ { 0.02 } } -\frac { 1 }{ { (-0.1) } } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left| { from\, lens\, formula:\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f }  } \right.  \ \, \, \, \, \, \, \, \Rightarrow 50+10=60\, m \ \therefore \, \, Power\, \, of\, the\, lens:\, \, P=\frac { 1 }{ f } =60\, D \ Now,\, \, \, consider\, the\, far\, point\, is\, \, 100\, cm. \ where, \ \, u=-100cm=-1m,\, \, \, and\, \, \, \, V=2\, cm=0.02\, m \ then, \ \frac { 1 }{ f } =\frac { 1 }{ { 0.02 } } -\frac { 1 }{ { (-1) } } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { we\, have\, the\, lens\, formla: } \right. \frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f }  \ \frac { 1 }{ f } =50+1=51\, m \ \therefore \, \, \, Power\, \, of\, the\, lens:\, \, P=\frac { 1 }{ f } =51\, D \ so,\, that\, the\, rangeof\, the\, \, power\, of\, eye\, lens\, is\, from\, \, \, \underline { 60\, D\, \, to\, \, 51\, D }  \ and\, the\, correct\, option\, is\, B. \ \, \,  \ so\, the\, correct\, optuon\, is\, A. \end{array}

The hyperopic eye can be corrected using --- lens.

structure of human eye human eye and colourful world

  1. concave

  2. convex

  3. plano convex

  4. plano concave

Show answer
Correct Option: B
Explanation:

Hyperopia or Far-sightedness is the condition where person can see the far objects clearly but cannot focus on near objects. This occurs since the image of nearby objects is formed behind retina. This corrected by using a convex lens which increases the power of the eye, and image is obtained on the retina.

Which type of lens is used to treat myopia:

structure of human eye human eye and colourful world

  1. Concave lens

  2. Convex lens

  3. Cylindrical lens

  4. Bifocal lens

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Correct Option: A
Explanation:

Near sightedness, or myopia, is the most common refractive error of the eye. Myopia occurs when the eyeball is too long, relative to the focusing power of the cornea and lens of the eye. This causes light rays to focus at a point in front of the retina, rather than directly on its surface. So, to treat myopia the glasses will be preceded by a minus sign ().
Hence, to treat myopia concave lens should be used.