Tag: human eye and colourful world

Questions Related to human eye and colourful world

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 $64-40=24D$.

A person cannot see the objects clearly placed at distance more than 40 cm. He is advised to use lens of power:

structure of human eye human eye and colourful world

  1. -2.5D

  2. +2.5D

  3. -6.25D

  4. +1.5D

Show answer
Correct Option: A
Explanation:

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).

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:

structure of human eye human eye and colourful world

  1. +50 D

  2. -50 D

  3. +2 D

  4. -2 D

Show answer
Correct Option: D
Explanation:

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)

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

Match the items in list-I with items in list-II and collect the correct answers from the codes given below the lists:

List-I List-II
I. Myopia A. Bifocal lens
II. Hyper-metropia B. Cylindrical lens
III. Presbyopia C. Concave lens
IV. Astigmation D. Convex lens
structure of human eye human eye and colourful world

  1. I-D, II-C, III-A, IV-B

  2. I-C,11-D,III-A, IV-B

  3. I-B, II-D, III-A, IV-C

  4. I-A, II-B, III-C, IV-D

Show answer
Correct Option: B
Explanation:

Myopia - Diverging lens (Concave)

Hyper-metropia - Converging lens (convex0
Presbyopia - Bifocal lenses
Astigmation - Cylindrical lenses.
thus,
I - C
II - D
III - A
IV - B

Option B is correct.

A person can see clearly objects between 15 and 100 cm from his eye. The range of his vision if he wears close fitting spectacles having a power of 0.8 diopter is :

structure of human eye human eye and colourful world

  1.  5 to 500 cm

  2.  12 to 250 cm

  3.  17 to 500 cm

  4.  17 to 250 cm

Show answer
Correct Option: C
Explanation:

Given, $v= \infty $

Focal length, $f= \dfrac{1}{p}$

$= \dfrac{100}{0.8}$

$= 125 cm$

To see the objects before 15 cm 


$v= +15\ ,\ u=?$

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

$\dfrac{1}{15}-\dfrac{1}{u}= \dfrac{1}{125}$

$u= 17.04\ cm$

To see the object far away from 100 cm

$u= 100cm,\ v= ?$

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

$\dfrac{1}{v}-\dfrac{1}{100}= \dfrac{1}{125}$

$v= \ 500cm$

A person can see clearly object only when they lie between 50 cm and 400 cm from his eyes. In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens, the person has to use, will be :

structure of human eye human eye and colourful world

  1. Convex, + 0.15 D

  2. Convex, +2.25 D

  3. Concave, -0.25 D

  4. Convex, + 0.2 D

Show answer
Correct Option: C
Explanation:

Image distance, $v=-4\ m$


We know that: $\dfrac{1}{-4}-\dfrac{1}{\infty}=\dfrac{1}{d}$

$\Rightarrow P=\dfrac{1}{d}=-0.25 $ D

 It is a concave lens.

A far sighted person has his near point $50$cm, find the power of lens he should use to see at $25$cm, clearly.

structure of human eye human eye and colourful world

  1. $+1$D

  2. $+2$D

  3. $-2$D

  4. $-1$D

Show answer
Correct Option: B
Explanation:

Given: A far sighted person has his near point 50cm, 

To find the power of lens he should use to see at 25cm, clearly.
Solution:
Distant objects need to be imaged at most 50 cm from the eye.
According to the given criteria,
$u=25cm, v=-50cm$
Applying lens formula, we get
$\dfrac 1f=\dfrac 1v+\dfrac 1u\\implies \dfrac 1f=\dfrac 1{-50}+\dfrac 1{25}\\implies \dfrac 1f=\dfrac {-1+2}{50}\\implies f=50cm=0.5m$
The power of the lens he should use is,
$P=\dfrac 1f=\dfrac 1{0.5}=+2D$

To read a poster on a wall, a person with defective vision needs to stand at a distance of $0.4m$ from the poster. A person with normal vision can read the poster from a distance of $2.0m$. Which one of the following lens may be used to correct the defective vision?

structure of human eye human eye and colourful world

  1. A concave lens of $0.5D$

  2. A concave lens of $1.0D$

  3. A concave lens of $2.0D$

  4. A convex lens of $2.0D$

Show answer
Correct Option: C
Explanation:

$u=-2m;v=-0.4m$
$\cfrac { 1 }{ f } =\cfrac { 1 }{ v } -\cfrac { 1 }{ u } $
$P=\cfrac { 1 }{ f } =-\cfrac { 1 }{ 0.4 } -\cfrac { 1 }{ (-2) } =-2$
$P=-2D$
that means concave lens of power $2D$

The nearer point of hypermetropic eye is 40 cm. The lens to be used for its correction should have the power?

structure of human eye human eye and colourful world
    • 1.5 D
    • 1.5 D
    • 2.5 D
    • 0.5 D
Show answer
Correct Option: C
Explanation:
Hypermetropia is corrected by using convex lens.
Focal length of lens used f = +(defected near point)
f = +d = +40cm
$\therefore$ power of lens = $\dfrac{100}{f(cm)} = \dfrac{100}{+40} = +2.5D$

The human eye has an approximate angular resolution of $\phi = 5.8 \times 10^{-4}$rad and typical photoprinter prints a minimum of 300 dpi (dots per inch, 1 inch = 2.54 cm). At what minimal distance z should a printed page be held so that one does not see the individual dots? 

structure of human eye human eye and colourful world

  1. 14.5 cm

  2. 20.5 cm

  3. 29.5 cm

  4. 28 cm

Show answer
Correct Option: A
Explanation:

Here, angular resolution of human eye,
$\phi \, = \, 5.8 \, \times \, 10^{-4} \, red$
The linear distance between two successive dots in a typical photo printer is $l \, = \, \dfrac{2.54}{300} \, cm \, = \, 0.84 \, \times \, 10^{-2} \, cm.$
At a distance of z cm, the gap distance l will subtend an angle
$\phi \, = \, \dfrac{l}{z} \, \therefore \, z \, = \, \dfrac{l}{\phi} \, = \, \dfrac{0.84 \, \times \, 10^{-2} \, cm}{5.8 \, \times \, 10^{-4}} \, = \, 14.5 \, cm$