Tag: structure of human eye

Questions Related to structure of human eye

A man's near point is 0.5 m and far point is 3 m. Power of spectacle lenses required for (i) reading purposes, (ii) seeing distant objects, respectively, are

structure of human eye human eye and colourful world
    • 2 D and + 3 D
    • 2 D and - 3 D
    • 2 D and - 0.33 D
    • 2 D and + 0.33 D
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Correct Option: C
Explanation:
For reading purposes:

$u = - 25 cm,   v = - 50  cm,  f = ?$

$\displaystyle \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} = - \dfrac{1}{50} + \dfrac{1}{25} = \dfrac{1}{50} ;         P = \dfrac{100}{f} = + 2 D$

For distant vision,  $ f' = \text{distance of far point} = - 3m$

$P = \displaystyle \dfrac{1}{f'} = - \dfrac{1}{3} D = - 0.33 D$

A person can see clearly between 1 m and 2 m his corrective lenses should be 

structure of human eye human eye and colourful world

  1. Bifocals with power -0.5 D and additional +3.5 D

  2. bifocals with power -1.0 D and additional +3 D

  3. Concave with power 1.0 D

  4. convex with power 0.5 D

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Correct Option: A

Which of the following lenses of appropriate focal length is used to correct shortsightedness ?

structure of human eye human eye and colourful world

  1. Concave

  2. Convex

  3. Both A & B

  4. None of the above

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Correct Option: A
Explanation:

A convex lens of appropriate focal length is used to correct farsightedness.
Farsightedness is the inability to see nearby objects clearly due to the formation of image by convergence of rays at a point behind the retina. A convex or converging lens with the correct focal length allows the rays to converge at a point on the retina. The convex lens first converge the rays before it reaches the eye and thus, reducing the image distance. The eye-lens will now be able to converge the refracted rays ton the retina.

A person who can see near objects clearly but cannot see clearly the distant objects is called hyperopia

structure of human eye human eye and colourful world

  1. True

  2. False

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Correct Option: B
Explanation:

Short sighted is also known as myopia, in which person cannot see distant objects clearly but able to see the near object clearly,  this also leads to blurred distance vision and is a very common problem which can be corrected by glasses or contact lenses.

A person who can see distant objects clearly but cannot see near objects clearly is said to be myopia

structure of human eye human eye and colourful world

  1. True

  2. False

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Correct Option: B
Explanation:
Long slighted, also know as hyperopia person ability to see objects close to them, these Vision problems are also referred as refractive errors.

In this, a person usually can see a distant object clearly but nearby objects will be out of focus.

Fill in the blacks with the correct words to make true statements:

Near sightendness and far sightedness happen when light is not focussed on the ...............

structure of human eye human eye and colourful world

  1. retina

  2. cornea

  3. pupil

  4. lens

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Correct Option: A
Explanation:

Nearsightedness results in blurred vision when the visual image is focused in front of the retina, rather than directly on it.


Farsightedness is the result of the visual image being focused behind the retina rather than directly on it.

Aperture of the human eye is $2 mm$. Assuming the mean wavelength of light to be $5000 \mathring{A}$ the angular resolution limit of the eye is:

structure of human eye human eye and colourful world

  1. $2 min$

  2. $1 min$

  3. $0.5 min$

  4. $1.5 min$

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Correct Option: B
Explanation:
Answer is B.
The angular resolution limit of eye = $\displaystyle = \frac{1.22\lambda}{a}$
$\displaystyle \frac{1.22 \times 5 \times 10^{-7}}{2 \times 10^{-3}}$
$\displaystyle \frac{1.22 \times 5 \times 10^{-7}}{2 \times 10^{-3}} \times \frac{180}{\pi} \times 60$ minutes
= 1 minute.
Hence, the angular resolution limit of the eye is $1 minute$.

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.

structure of human eye human eye and colourful world

  1. 0.25 D

  2. -0.25 D

  3. 0.5 D

    • 0.75 D
Show answer
Correct Option: B
Explanation:

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.

The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect?

structure of human eye human eye and colourful world
    • 1.0D

  2. -1.0D

  3. +3.0D

  4. -3.0D

Show answer
Correct Option: C
Explanation:

$Answer:-$ C option

In the problem, it is given that the near point of defective eye is 1 m and that of a normal eye is 25 cm. Hence $u = -25$ cm. The lens used forms its virtual image at near point of hypermetropic eye i.e., $v = - 1 m = -100$ cm. 

Using lens formula, we have :-

$\dfrac { 1 }{ v } -\dfrac { 1 }{ u } =\dfrac { 1 }{ f } \\ \dfrac { 1 }{ -100 } -\dfrac { 1 }{ -25 } =\dfrac { 1 }{ f } \\ f=\dfrac { 100 }{ 3 } =0.33\quad m\\ power=\dfrac { 1 }{ f(in\quad metres) } =+3.0\quad D$

For a normal eye, the far point is at infinity and the near point of distinct vision is about $25\ cm$ in front of the eye. The cornea of the eye provides a converging power of about $40$ dioptres, and the least converging power of the eye - lens behind the cornea is about $20$ dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye.

structure of human eye human eye and colourful world

  1. $10$ to $14\ D$

  2. $20$ to $24\ D$

  3. $28$ to $32\ D$

  4. $14$ to $18\ D$

Show answer
Correct Option: A
Explanation:

To see object at infinity, eye uses its least converging power.

Power of eye lens, $P=40+20=60D$
Power of eye lens is $1/f\Rightarrow f=5/3 cm$
To focus on object at the near point, object distance $u=-d=-25$ cm
Focal length of eye lens is distance between the cornea and the retina.
Image distance, $v=5/3\ cm$
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f _1}\Rightarrow f _1=16/25cm^{-1}$
Power is $100/f _1=64D$
power of eye lens is $64-40=24D$.