Physics

Optical Instruments and Human Eye

416 Questions

Optical instruments and the human eye explore how lenses and mirrors are used to magnify and resolve images. Key topics include the magnifying power of astronomical telescopes, compound microscope configurations, and hyperfocal distance calculations. These physics concepts are vital for general science competitive exams.

Telescope magnifying powerCompound microscope lensesHyperfocal distance calculationHuman eye resolutionSpherical mirror magnification

Optical Instruments and Human Eye Questions

Multiple choice structure of human eye human eye and colourful world

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

    • 1.0D

  2. -1.0D

  3. +3.0D

  4. -3.0D
Reveal answer Fill a bubble to check yourself
C Correct answer
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 :-

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

Multiple choice structure of human eye 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.

  1. $10$ to $14\ D$
  2. $20$ to $24\ D$
  3. $28$ to $32\ D$
  4. $14$ to $18\ D$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $64-40=24D$.

Multiple choice structure of human eye human eye and colourful world

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

  1. -2.5D

  2. +2.5D

  3. -6.25D

  4. +1.5D
Reveal answer Fill a bubble to check yourself
A Correct answer
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).

Multiple choice structure of human eye human eye and colourful world

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:

  1. +50 D

  2. -50 D

  3. +2 D

  4. -2 D
Reveal answer Fill a bubble to check yourself
D Correct answer
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)

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

Multiple choice structure of human eye human eye and colourful world

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 :

  1. 5 to 500 cm

  2. 12 to 250 cm

  3. 17 to 500 cm

  4. 17 to 250 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
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

$u= 100cm,\ v= ?$

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

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

$v= \ 500cm$

Multiple choice structure of human eye human eye and colourful world

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 :

  1. Convex, + 0.15 D

  2. Convex, +2.25 D

  3. Concave, -0.25 D

  4. Convex, + 0.2 D
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice structure of human eye human eye and colourful world

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

  1. $+1$D
  2. $+2$D
  3. $-2$D
  4. $-1$D
Reveal answer Fill a bubble to check yourself
B Correct answer
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,
$P=\dfrac 1f=\dfrac 1{0.5}=+2D$

Multiple choice structure of human eye human eye and colourful world

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?

  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 $2.0D$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $2D$

Multiple choice structure of human eye human eye and colourful world

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

    • 1.5 D
    • 1.5 D
    • 2.5 D
    • 0.5 D
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Hypermetropia is corrected by using convex lens.
Focal length of lens used f = +(defected near point)
f = +d = +40cm
$\therefore$ power of lens = $\dfrac{100}{f(cm)} = \dfrac{100}{+40} = +2.5D$
Multiple choice structure of human eye human eye and colourful world

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? 

  1. 14.5 cm

  2. 20.5 cm

  3. 29.5 cm

  4. 28 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
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
$\phi \, = \, \dfrac{l}{z} \, \therefore \, z \, = \, \dfrac{l}{\phi} \, = \, \dfrac{0.84 \, \times \, 10^{-2} \, cm}{5.8 \, \times \, 10^{-4}} \, = \, 14.5 \, cm$

Multiple choice structure of human eye human eye and colourful world

For a normal eye, the cornea of eye provides as converging power of 40 D and the lens least converging power of the eye lens behind the cornea is 20 D. Using this information , the distance between the retina and the cornea – eye lens can be estimated to be:

  1. 1.5 cm

  2. 5 cm

  3. 2.5 cm

  4. 1.67 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given that,

Power ${{P} _{1}}=40\,D$

Power ${{P} _{2}}=20\,D$

Total power of the combination

  $ P={{P} _{1}}+{{P} _{2}} $

 $ P=40+20 $

 $ P=60\,D $

Now, focal length of the combination

  $ f=\dfrac{1}{60}\times 100 $

 $ f=\dfrac{5}{3}\,cm $

Now, for minimum converging state of the eye lens object is at infinity

  $ u=-\infty  $

 $ f=\dfrac{5}{3} $

Now, from lens equation

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

 $ \dfrac{3}{5}=\dfrac{1}{v}-\dfrac{1}{\infty } $

 $ \dfrac{1}{v}=\dfrac{3}{5} $

 $ v=\dfrac{5}{3}\,cm $

So, the distance between retina and cornea – lens system $1.67\ cm$

Multiple choice physics the human eye and the colourful world human eye and its working what is inside our eyes? human eye and defects of vision

The changing of focal length of an eye-lens to focus the image of an object at varying distances is done by the action of the :

  1. pupil

  2. ciliary muscles

  3. retina

  4. blind spot
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

cilliary muscles,contraction and relaxation changes the focal length of eye lens to focus the image of an object at varying distance.

Multiple choice physics the human eye and the colourful world human eye and its working what is inside our eyes? human eye and defects of vision

Two pont white are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm. Approximately, what is the maximum distance at which these dots can be resolved by the eye ? [Take wavelength of light = 500 nm]

  1. 5 m

  2. 1 m

  3. 6 m

  4. 3 m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Rayleigh's criterion, the angular resolution is theta = 1.22 * lambda / D. With lambda = 500 nm and D = 3 mm, theta = 1.22 * 500e-9 / 0.003 = 2.03e-4 radians. For a separation of 1 mm (0.001 m), the distance L = 0.001 / 2.03e-4, which is approximately 4.92 meters, rounding to 5 meters.

Multiple choice physics the human eye and the colourful world human eye and its working what is inside our eyes? human eye and defects of vision

The distance of the eye-lens from the retina is $x$. For a normal eye, the maximum focal length of the eye-lens

  1. $= x$
  2. $< x$
  3. $> x$
  4. $= 2x$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Focal length of normal eye lens is equal to the distance between eye-lens and retina i.e, $x$  so that all rays converges on retina after passing through  eye-lens