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 physics observing space: telescopes space research and satellites space travel space exploration and forms of light

In an astronomical telescope, the distance between the objective and the eyepiece is 36 cm and the final image is formed at infinity. The focal length $f _0$ of the objective and the focal length $f _e$ of the eyepiece are

  1. $f _0=45 cm$ and $f _e=-9 cm$
  2. $f _0=50 cm$ and $f _e=22 cm$
  3. $f _0=65 cm$ and $f _e=7 cm$
  4. $f _0=30 cm$ and $f _e=6 cm$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

The focal lengths be $f _{o}$ and $f _{e}$ 


then $f _{o}+f _{e}=36$ for the image to be at infinity 

option $A$ is correct as $45-9=36$

Multiple choice physics observing space: telescopes space research and satellites space travel space exploration and forms of light

The focal length of eye lens and object lens of a telescope is 4 mm and 4 cm respectively. If final image of an far object is at $\displaystyle \infty $. Then the magnifying power and length of the tube are:

  1. 10, 4.4 cm

  2. 4, 44 cm

  3. 44, 10 cm

  4. 10, 44 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Magnification is the amount that a telescope enlarges its subject. Its equal to the telescopes focal length divided by the eyepieces focal length. As a rule of thumb, a telescopes maximum useful magnification is 50 times its aperture in inches (or twice its aperture in millimeters). 
That is, $M = fo / fe$
In this case, the focal length of eye lens and object lens of a telescope is 4 mm = 0.4 cm and 4 cm respectively.
So, Magnification $M = fo / fe = 4 / 0.4 = 10.$
Focal length of the eyepiece is the distance from the center of the eyepiece lens to the point at which light passing through the lens is brought to a focus.
Focal length of the objective is the distance from the center of the objective lens (or mirror) to the point at which incoming light is brought to a focus.
The length of the tube is given as sum of the focal lengths of the eye lens and the object lens.
So, Length of the tube $ = fo + fe = 4 + 0.4 = 4.4 cm.$
Hence, the magnifying power and length of the tube are: 10, 4.4 cm.

Multiple choice physics observing space: telescopes space research and satellites space travel space exploration and forms of light

An astronomical telescope has an eyepiece of focal length $5 cm$. If magnification produced is $14$ in normal adjustment, then what is the length of telescope?

  1. $25 cm$
  2. $75 cm$
  3. $50 cm$
  4. $100 cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle m = \frac {f _o}{f _e}$ or $\displaystyle 14 = \frac {f _o}{5} \Rightarrow f _o = 70 cm$

$\displaystyle \therefore L=f _o + f _e = 5 + 70 = 75 cm$

Multiple choice physics observing space: telescopes space research and satellites space travel space exploration and forms of light

A planet is observed by an astronomical reflecting telescope having an objective of focal length 16 m and an eye - piece of focal length 2 cm.

  1. the distance between the objective and the eye - piece is 16.02 m.

  2. the angular magnification of the planet is 800.

  3. the image of planet is erect.

  4. the objective is larger than eye - piece.
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

Length of astronomical reflecting= distance between the objective and the eye - piece =L, 

$f _o=16m$ and $f _e=2cm=0.02m$
$L=f _o+f _e=16+0.02m=16.02m$
Angular magnification $m=\dfrac{f _o}{f _e}=\dfrac{16}{0.02}=800$. 
In astronomical reflecting telescope, the image formed is inverted.
The objective is always larger than eye - piece.

Multiple choice physics observing space: telescopes space research and satellites space travel space exploration and forms of light

The anguThe angular resolution of 10cm diameter telescope at a wavelength of $5000nm$ is of order of the

  1. $ 10^{4}$ rad
  2. $ 10^{-4}$ rad
  3. $ 10^{-6}$ rad
  4. $ 10^{6}$ rad
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given that the Diameter of telescope is $D=10cm=0.1m$ and the wavelength $\lambda =5000nm$

Angular resolution $d\theta =\frac{1.22 \lambda}{D},$
$\Rightarrow d\theta = \frac{1.22 \times 5000 \times 10^{-9}}{0.1}=0.61 \times 10^{-6},$
Therefore it is in the order of $10^{-6},$
So the correct option is $C.$

Multiple choice physics observing space: telescopes space research and satellites space travel space exploration and forms of light

The optical length of an astronomical telescope with magnifying power of 10, for normal vision is 44cm. What is focal length of the objective?

  1. 40cm

  2. 22cm

  3. 10cm

  4. 4cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Length of telescope $L = f _o + f _e$
$\therefore$ $f _o + f _e  = 44$
We get $f _e = 44 - f _o$
Magnification of astronomical telescope for normal vision $|M| = \dfrac{f _o}{f _e}$
OR $10 = \dfrac{f _o}{44 - f _o}$
OR $440 - 10 f _o = f _o$

$\implies$ $f _o = \dfrac{440}{11} =40$ cm

Multiple choice physics observing space: telescopes space research and satellites space travel space exploration and forms of light

The length of an astronomical telescope for normal vision is _______.

  1. $-\displaystyle\frac{f _o}{f _e}$
  2. $-f _o\times f _e$
  3. $\displaystyle\frac{f _e}{f _o}$
  4. $f _o+f _e$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

To get a magnified image of an object at infinity the image formed by objective lens will beat focus and if we want maximum magnification then the image formed by objective should be at focus of the eye piece.

Hence length of telescope =${ f } _{ o }+{ f } _{ e }$.
Here ${ f } _{ o }$ and ${ f } _{ e }$ are focal length of objective and eyepiece lens respectively.

Multiple choice physics observing space: telescopes space research and satellites space travel space exploration and forms of light

For the astronomical telescope, the focal length of objective lens is ${f} {o}$ and the eye piece lens is ${f} _{e}$. Then the tube length of the telescope is ____

  1. $L\ge { f } _{ o }-{ f } _{ e }$
  2. $L\ge { f } _{ o }+{ f } _{ e }$
  3. $L<{ f } _{ o }+{ f } _{ e }$
  4. $L\le { f } _{ o }-{ f } _{ e }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For astronomical telescope
$\left| { v } _{ 1 } \right| ={ f } _{ o }$
$\left| { u } _{ 2 } \right| \le { f } _{ e }$
Probable answer would be (c) conceptually correct

Multiple choice physics observing space: telescopes space research and satellites space travel space exploration and forms of light

The aperture of the largest telescope in the world is $5 m,$ if the separation between the Moon and the Earth is $4 \times 10^5 km$ and the wavelength of the visible light is $5000 \overset {o}{A}$ then the minimum separation between the objects on the surface of the Moon which can be just resolve is approximately

  1. $1 m$
  2. $10 m$
  3. $50 m$
  4. $200 m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics observing space: telescopes space research and satellites space travel space exploration and forms of light

If an astronomical telescope has objective and eye-pieces of focal length 200 cm and 4 cm respectively,then the magnifying power of the telescope for the normal vision is:

  1. 42

  2. 50

  3. 58

  4. 204
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

 

It is given that,

Focal length of eye- piece fe = 4 cm

Focal length of object is fo = 200 cm

Least distance of distinct vision is d = 25 cm

So, magnifying power of microscope is

$ M=\dfrac{-{{f} _{0}}}{{{f} _{e}}}\left( 1+\dfrac{{{f} _{e}}}{d} \right) $

$ M=\dfrac{-200}{4}\left( 1+\dfrac{4}{25} \right) $

$ =-58\,cm $