Physics

Optical Instruments and Human Eye

428 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 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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $250 cm$.

Multiple choice structure of human eye human eye and colourful world

The power of accommodation for the normal eye is

  1. 4 D

  2. 40 D

  3. 44 D

  4. 400 D
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Our normal vision length = 25 cm which is eye lens focal length focal length  $f = 0.25 m$

Power $P=\dfrac{1}{f}=\dfrac{1}{.25}=4D$

Multiple choice structure of human eye human eye and colourful world

The near point of a person is $75cm$. In order that he may be able to read book at a distance $25cm$. The power of spectacles lenses should be

  1. $-2D$
  2. $+3.75D$
  3. $+2.6D$
  4. $+3D$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the lens formula 1/f = 1/v - 1/u. Here, v = -75 cm and u = -25 cm. 1/f = 1/-75 - 1/-25 = -1/75 + 3/75 = 2/75. Power P = 100/f = 100 * (2/75) = 200/75 = 2.66D.

Multiple choice structure of human eye human eye and colourful world

An old person is able to see an object nearest to 1 m.What should be the power of lens required so that he can see an object placed at nearest distance of distinct vision?

  1. 3D

  2. 4D

  3. 1/2D

  4. 1/4D
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Hypermetropia is a condition of the eye where the person is not able to see things clearly when nearer to the eye.

The normal near point of the eye is \[u=25\text{ }cm\]

Focal length, $f$

Image formed at distance, $v=100\,cm$

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

$ \dfrac{1}{f}=\dfrac{1}{25}-\dfrac{1}{100}=\dfrac{3}{100} $

$ f=33.33\,cm=0.33\,m $

Power, $P=\dfrac{1}{f}=\,\dfrac{1}{0.333}=3D$

Hence, Power of lens is $3D$

Multiple choice structure of human eye human eye and colourful world

A far-sighted person cannot focus distinctly objects closer than 120 cm. The lens that will permit him to read from a distance of 40 cm will have a focal length:

    • 30 cm
    • 30 cm
    • 60 cm
    • 60 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,

Near point, $v=-120\,cm$

Reading Distance, $u=-40\,cm$

From lens formula,

  $ \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{-120}-\dfrac{1}{-40}=\dfrac{1}{60} $

 $ f=+60\,cm $

Focal length of lens $f=+60\,cm$ 

Multiple choice structure of human eye human eye and colourful world

For a normal eye, The cornea of eye provides a converging power of $40 \,D$ and the 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
Power of Cornea, $P _1=40 D$
and power of the eye lens,  $P _2 =20D$
${ P } _{ eff }= P _1+P2$
$\Rightarrow P _{eff}=40D+20D=60D$
So, $f=\dfrac{100}{P _{ eff }}=\dfrac{100}{60}=1.67cm$
Hence, the correct option is $(D)$

Multiple choice structure of human eye human eye and colourful world

Which of the following is true for a person suffering from myopia?

  1. Can see far-off object clearly but near objects appear blurred

  2. can be corrected using a convex lens

  3. far point is at finite distance, and not at infinity

  4. near point is beyond 25 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In myopia, the eyeball is too long or the lens is too curved, causing light to focus in front of the retina. Consequently, the far point is brought closer than infinity, making distant objects blurry.

Multiple choice structure of human eye human eye and colourful world

A person can see clearly objects 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, $+2.25$ dioptre
  2. Concave, $-0.25$ dioptre
  3. Concave, $-0.2$ dioptre
  4. Convex, $+0.15$ dioptre
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Maximum distance of distinct vision $=400$cm. So image of object at infinity is to be formed at $400$cm


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

$\dfrac{1}{-400}-\dfrac{1}{\infty}=\dfrac{1}{f}$

$P=-0.25$D.

Multiple choice structure of human eye human eye and colourful world

A far sighted person can see object beyond $71\;cm$ clearly if the separation between the glasses and eye lens is $2\;cm$, then find the focal length of glasses.

  1. $23\;cm$
  2. $34.5\;cm$
  3. $18.4\;cm$
  4. $30\;cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a farsighted person, the lens must form a virtual image of an object at the near point (71 cm) at the person's actual near point. Accounting for the 2 cm separation between the lens and eye, the object distance u is infinity and image distance v is -(71 - 2) = -69 cm, or using standard lens formula with u = -71 cm and v = -(71 - 2) = -69 cm. Using the lens formula 1/f = 1/v - 1/u yields the correct focal length.

Multiple choice structure of human eye human eye and colourful world

A person suffering from eyesight defect has far point at $40cm$ and near point at $25cm$. The person uses a lens to see far away object. Find the near point of the person while wearing this lens.

  1. $50cm$
  2. $\dfrac{200}{13}cm$
  3. $\dfrac{200}{3}cm$
  4. $25cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice structure of human eye human eye and colourful world

A farsighted woman breaks her current eyeglasses and is using an old pair whose refractive power is 1.660 diopters. Since these eyeglasses do not completely correct her vision, she must hold a newspaper 42.00 cm from her eyes in order to read it. She wears the eyeglasses 2.00 cm from her eyes. How far is her near point from her eyes?

  1. 75 cm

  2. 125 cm

  3. 225 cm

  4. 121.05 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The refractive power of the lens is $1.660$ dipters. So the focal length of the lens is $f=\dfrac{1}{1.660}=0.6024\,m=60.24\,cm$

The distance between the newpaper and her eyes is $42.00\,cm$

The distance between her eyes and glasses is $2.00\,cm$

So, the distance between the lglasses and newspaper is $(42.00-2.00)\,cm=40.00\,cm.$ That is, $d _0=40.00\,cm$

The distance $(d _i)$ between the glasses and the virtual image formed by the lens given by,

$d _i=\dfrac{fd _0}{d _0-f}=\dfrac{60.24-40.00}{40.00-60.24}=-119.05\,cm$

This relation is obtained from thin lens equation and the negative sign implies the image is virtual.

This position is her near point.

So the near point from her eyes is, $119.05\,cm+2.00\,cm=121.05\,cm$
Multiple choice structure of human eye human eye and colourful world

A man can see clearly up to $3\ \textit{metres}$. Prescribe a lens for his spectacles so that he can see clearly up to $12\ \textit{metres}$

  1. $-3/4\ D$
  2. $3\ D$
  3. $-1/4\ D$
  4. $-4\ D$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The lens must form an image at 3 m for an object at 12 m. 1/f = 1/v - 1/u = 1/-3 - 1/-12 = -4/12 + 1/12 = -3/12 = -1/4. Power P = 1/f = -0.25 D.

Multiple choice structure of human eye human eye and colourful world

A person cannot see distinctly at the distance less than one metre. Calculate the power of the lens that he should use to read a book at a distance of $25\, cm$

  1. $+ 3.0\, D$
  2. $+ 0.125\, D$
  3. $- 3.0\, D$
  4. $+ 4.0\, D$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The person needs to see an object at 25 cm (u = -0.25 m) as if it were at 1 m (v = -1 m). 1/f = 1/v - 1/u = 1/-1 - 1/-0.25 = -1 + 4 = 3. Power P = +3.0 D.

Multiple choice structure of human eye human eye and colourful world

A person has a defect of eye-vision his near point is at distance of 40 cm from his eye it means 

  1. he cannot observe the objects clearly which are at a distance more than 40 cm

  2. he can observe the objects clearly which are at a distance of 40 cm only.

  3. he can observe the objects clearly which are at a distance equal to 40 cm or more than 40 cm

  4. he can observe the objects clearly which hare at a distance less than 40 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A near point of 40 cm means the person's eye can clearly focus on objects placed at 40 cm and farther away, but struggles with objects closer than 40 cm. Therefore, they can observe objects clearly at a distance equal to 40 cm or more than 40 cm. Option C accurately captures this definition of hypermetropia.

Multiple choice structure of human eye human eye and colourful world

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

    • 2 D and + 3 D
    • 2 D and - 3 D
    • 2 D and - 0.33 D
    • 2 D and + 0.33 D
Reveal answer Fill a bubble to check yourself
C Correct answer
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$

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