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 option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

A telescope has an objective of focal length $50 cm$ and an eye-piece of focal length $5 m$ THe least distance of distinct vision is $25 cm$. The telescope is focused for distinct vission a scale $200 cm$ away from the objective.The separation between the two lenses is nearly 

  1. $71 cm$
  2. $61 cm$
  3. $81 cm$
  4. $51 cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics observing space: telescopes optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

The sum of the focal lengths of the objective and an eyepiece, In case of an astronomical telescope is equal to : ( final image is at $\infty$)

  1. The length of the telescope

  2. Half the length of the telescope

  3. Double the length of the telescope

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Answer is A.

The astronomical telescope makes use of two positive lenses: the objective, which forms the image of a distant object at its focal length, and the eyepiece, which acts as a simple magnifier with which to view the image formed by the objective. Its length is equal to the sum of the focal lengths of the objective and eyepiece, and its angular magnification is -fo/fe , giving an inverted image.
Hence, the sum of the focal lengths of the objective and an eyepiece, In case of an astronomical telescope is equal to the length of the telescope.

Multiple choice physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

The focal lengths of the objective and the eyepiece of an astronomical telescope are 20 cm and 5 cm respectively. If the final image is formed at a distance of 30 cm from the eyepiece, find the separation between the lenses required for distinct vision

  1. 32.4 cm

  2. 42.3 cm

  3. 24.3 cm

  4. 30.24 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$f _{0}=20cm$


$f _{e}=5cm$

$V _{e}=30cm = D$

$L=?$

$L _{D}=f _{0}+ \dfrac{Df _e}{D+f _e}$ $=20+\dfrac{5\times 30}{35}=24.3 cm$

Multiple choice physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

The focal length of the objective of an astronomical telescope is 1 m and it is in normal adjustment.Initially the telescope is focussed to a heavenly body. If the same telescope is to be focussed to an object at a distance of 21 m from the objective,then identify the correct choice

  1. eye piece should be displaced by 2 cm away from the objective

  2. eye piece should be displaced by 2 cm towards the objective

  3. eye piece should be displaced by 5 cm towards the objective

  4. eye piece should be displaced by 5 cm away from the objective
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know, $ \dfrac {1}{f} = \dfrac {1}{v} - \dfrac {1} {u} $


We seeing heavenly body,  $u =  \infty$
$v = f = 1 m$

When seeing 21 m far
$u = - 21 m $
$f = 1 m$
$v = \dfrac {f u} {(f+u)} = 21 / 20 = 1.05 m$

So, eye piece need to move $1.05-1 = 0.05 m$ further away from objective

Answer. D) eye piece should be displaced by $5 cm$ away from the objective

Multiple choice physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

The focal lengths of the objective and eyepiece of a telescope are 60cm and 5cm respectively.The telescope is focused on an object 360cm from the objective and the final image is formed at a distance of 30cm from the eye of the observer. The length of the telescope is

  1. 66.3 cm

  2. 86.3 cm

  3. 76.3 cm

  4. 96.3 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$f _o = 60cm$


$f _e = 5cm$

$r _e = 30cm$

$L = ?$

$u _o = 360cm$

$\dfrac {1}{v _o} + \dfrac {1}{360} = \dfrac {-1}{60}$

$\dfrac {1}{v _o} = \dfrac {-1}{60} + \dfrac {1}{360}$

Multiple choice physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

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

  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

A telescope uses two co-axially placed convex lenses in such a way that the focus of objective lens is past the focus of the eye piece as evident here from the focal lengths of the objective lens and the eye piece. $ \therefore $ The objective lens is larger than eye piece.

Length of the tube is given as 
$ L = f _o + f _e = 16\ m + 2\ cm = 16.02\ m $

Angular magnification is given as:
$ m = \dfrac{f _o}{f _e} = \dfrac{1600}{2} = 800 $


Since, the first lens or the objective lens produces a real and inverted image of the object to be observed and this real image formed acts as an object for the eye piece convex lens and is between the focus and the optical center, so, an inverted, virtual and enlarged image of the object is formed. $ \therefore $ The final image is inverted. 

Hence, the correct answers are OPTIONS A,B and D.

Multiple choice physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of

  1. large focal length and small diameter

  2. large focal length and large diameter

  3. small focal length and large diameter

  4. small focal length and small diameter
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Angular magnification is proportional to the focal length of the objective, while angular resolution is proportional to the diameter of the objective. Therefore, both a large focal length and a large diameter are required.

Multiple choice physics universe and space science xx radio telescope and space telescopes launching vehicles optical instruments : telescope

The Corrective Optics Space Telescope Axial Replacement (COSTAR) system was designed to correct the _______________________ for light focused at the FOC, FOS, and GHRS.

  1. visual aberration

  2. spherical aberration

  3. diffraction

  4. none of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The Corrective Optics Space Telescope Axial Replacement(COSTAR) system was designed to correct the spherical aberration for light focused at the FOC, FOS and GHRS. Spherical aberration is the condition when there is a loss in the definition of the image arising from the surface geometry of a spherical mirror or lens. Hence to solve this  condition COSTAR was designed.

Multiple choice image formation by lens image formation by lenses ray optics and optical instruments optics physics

The lateral magnification of the lens with an object located at two different positions $u _1$ and $u _2$ are $m _1$ and $m _2$, respectively. Then the focal length of the lens is :

  1. $f=\sqrt {m _1m _2}(u _2-u _1)$
  2. $\dfrac{m _2u _2 - m _1u _1}{m _2-m _1}$
  3. $\dfrac {(u _2-u _1)}{\sqrt {m _2m _1}}$
  4. $\dfrac {(u _2-u _1)}{(m _2)^{-1}-(m _1)^{-1}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

Magnification is: $\dfrac{f}{u-f}$
$\dfrac{m _{1}}{m _{2}}=\dfrac{\frac{f}{u _{1}-f}}{\dfrac{f}{u _{2}-f}}$

$f=\dfrac{u _{2}m _{2}-u _{1}m _{1}}{m _{2}-m _{1}}$

Multiple choice physics observing space: telescopes xx radio telescope and space telescopes optical instruments : telescope

An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the observer the tree appears:

  1. 10 times taller .

  2. 10 times nearer .

  3. 20 times taller .

  4. 20 times nearer .
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Magnifying power of a telescope relates to the angular magnification. It makes the object appear closer by a factor equal to the magnifying power, effectively reducing the perceived distance.

Multiple choice physics option c: imaging xx radio telescope and space telescopes optical instruments : telescope

The focal length of objective and eye-piece of a telescope are $200\ cm$ and $5\ cm$ respectively. Final image is formed at least distance of distinct vision. The magnification of telescope is

  1. $20$
  2. $24$
  3. $30$
  4. $48$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a telescope where the final image is at the least distance of distinct vision (D = 25 cm), the magnification M = f_o / f_e * (1 + f_e / D). Plugging in f_o = 200, f_e = 5, and D = 25 gives M = (200/5) * (1 + 5/25) = 40 * 1.2 = 48.

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

The linear magnification for a mirror is the ratio of the size of the image to the size of the object, and is denoted by m. Then m is equal to (symbols have their usual meanings).

  1. $\displaystyle \frac { uf }{ u-f } $
  2. $\displaystyle \frac { uf }{ u+f } $
  3. $\displaystyle \frac { f }{ u-f } $

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

we now,$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$
multiplying by u in above eq.
$\dfrac{u}{f}=\dfrac{u}{v}+\dfrac{u}{u}$
$\dfrac{u}{f}=\dfrac{u}{v}+1$
$\dfrac{u}{f}-1=\dfrac{u}{v}$
$\dfrac{u}{v}=\dfrac{u-f}{f}$
$\dfrac{v}{u}=\dfrac{f}{u-f}  ,  As, m=\dfrac{v}{u}$
$m=\dfrac{f}{u-f}$
hence,option C is correct.