Tag: human eye and colourful world

Questions Related to human eye and colourful world

A person cannot see objects clearly beyond 50cm. The power of lens to correct the vision is

  1. +5D

  2. -2D

  3. -0.5D

  4. +2D


Correct Option: B
Explanation:

A person cannot see beyond 50 cm. Hence, he should use a concave lens of 50 cm focal length.Hence $Power=\dfrac{-100}{50} =-2 D$

Long-sighted people who have lost their spectacles can still read a book by looking through a small (3-4 mm) hole in a sheet of paper :

  1. Because the fine hole produces an image of the letters at a longer distance

  2. Because in doing so the distance of the object is increased

  3. Because in doing so the focal length of the eye lens is effectively decreased

  4. Because in doing so the focal length of the eye lens is effectively increased


Correct Option: C
Explanation:

Long-sighted people who have lost there spectacles can still read a book by looking through a small (3-4 mm) hole in a sheet of paper because by doing so due to the diffraction at the hole the focal length of the eye lens is effectively decreased. Hence correct option is C.

A man cannot see clearly the objects beyond a distance of 20 cm from his eyes. To see distant objects clearly the kind of lenses and its focal length must be 

  1. 100 cm convex

  2. 100 cm concave

  3. 20 cm convex

  4. 20 cm concave


Correct Option: D
Explanation:

$u= +20cm$

$v= \infty $

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

$\dfrac{1}{\infty }-\dfrac{1}{+20}= \dfrac{1}{f}$

$f=-20\ cm$

The power of a lens used to remove the myopic defect of an eye is 0.66 D. The far point for this eye is (nearly) :

  1. 25 cm

  2. 150 cm

  3. 100 cm

  4. 75 cm


Correct Option: B
Explanation:

As $p= \dfrac{1}{f}$


So, $f= \dfrac{1}{p}$

$\Longrightarrow$  $f= \dfrac{100}{0.66}cm$

$\Longrightarrow$ $f= 151.51cm$

$\Longrightarrow$ $f\simeq 150cm$

A myopic person can not see objects lying beyond 2m. The focal length and power of the lens required to remove this defect will be 

  1. 1m and 0.5D

  2. -2m and -0.5D

  3. 0.5m and 0.5D

  4. -0.5, and 0.5D


Correct Option: B
Explanation:

$u=+2m$

$v= \infty $

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

$\dfrac{1}{\infty }-\dfrac{1}{2}=\dfrac{1}{f}$

$f= -2m$

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

$= \dfrac{1}{-2}$

$= -0.5 D$

A long sighted person has a least distance of distinct vision of 50 cm. He wants to reduce it to 25 cm. He should use a :

  1. Concave lens of focal length 50 cm

  2. Convex of focal length 25 cm

  3. Convex lens of focal length 50 cm

  4. Concave lens of focal length 25 cm


Correct Option: C
Explanation:

Given, $u= \infty $


Image distance, $v= 50cm$

From lens formula, $\dfrac{1}{v}-\dfrac{1}{u}= \dfrac{1}{f}$

$\dfrac{1}{50}-\dfrac{1}{\infty }= \dfrac{1}{f}$

$f= +50\ cm$

He should use a convex lens of focal length 50 cm.

A person can see clearly objects lying between 25 cm and 2 m from his eye. His vision can be corrected by using spectacles of power:

  1. +0.25 D

  2. +0.5 D

  3. -0.25 D

  4. -0.5 D


Correct Option: D
Explanation:

Here, $u= 2m$
$v= \infty $

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

$\dfrac{1}{\infty}-\dfrac{1}{2}= \dfrac{1}{f}$

           $f= -2m$

Now, $P= \dfrac{1}{f}$

              $= \dfrac{1}{-2}D$

              $= -0.5\ D$

The power of lens, a short sighted person uses is 2 dioptre. The maximum distance of an object which he can see without spectacles is 

  1. 25cm

  2. 50cm

  3. 100cm

  4. 10cm


Correct Option: B
Explanation:

$f= \dfrac{1}{p}$

$= \dfrac{100}{2}cm$

$= 50\ cm$

So he can see upto a distance of 50 cm.

A person wears glasses of power 2D. The defect of the eye and the far point of the person without the glasses will be 

  1. Nearsighted, 50 cm

  2. Farsighted, 50cm

  3. Nearsighted, 250 cm

  4. Astigmatism, 50 m


Correct Option: B
Explanation:

$f= \dfrac{1}{p}$

  $=\dfrac{100}{2}cm$

  $=50\ cm$

The defect of eye will be farsighted.

The far point of a myopic eye is 250 cm. The correcting lens should be a

  1. diverging lens of focal length 250 cm

  2. converging lens of focal length 250 cm

  3. diverging lens of focal length 125 cm

  4. converging lens of focal length 125 cm


Correct Option: A
Explanation:

$u = -\infty $ 

$ v = -250 cm$

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

$(-\dfrac{1}{\infty}-\dfrac{1}{250})= \dfrac{1}{f}$

$f= -250cm$

The correcting lens of focal length $250 cm$.