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 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
C Correct answer
Explanation

The limit of resolution for a telescope is given by theta = 1.22 * lambda / D, where D is the aperture diameter and lambda is the wavelength. Multiplying this angular resolution by the distance to the moon gives the minimum linear separation, which calculates to approximately 50 meters.

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 $

 

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

The diameter of moon is $3.5\times{10}^{3}km$ and its distance from the earth is $3.8\times{10}^{5}km$. The focal length of the objective and eyepiece are $4m$ and $10cm$ respectively. The angle subtended by the diameter of the image of the moon will be approximately

  1. ${2}^{o}$
  2. ${20}^{o}$
  3. ${40}^{o}$
  4. ${50}^{o}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The angular diameter of the moon is alpha = diameter / distance = 3.5e3 / 3.8e5 radians. The telescope magnification M = f_o / f_e = 400 cm / 10 cm = 40. The image angle beta = M * alpha = 40 * (3.5/380) radians, which is approximately 0.368 radians, or about 21 degrees.

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

The magnifying power an astronomical telescope for normal adjustment is -

  1. $- \frac{f _0}{f _e}$
  2. $-f _0 \times f _e$
  3. $- \frac{f _e}{f _0}$
  4. $-f _0 + f _e$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The magnifying power of an astronomical telescope in normal adjustment is defined as the ratio of the focal length of the objective to that of the eyepiece, with a negative sign indicating an inverted image.