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 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 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

Linear magnification is

  1. Positive for an inverted image

  2. Negative for an erect image

  3. Zero for an inverted image

  4. Negative for an inverted image
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Linear magnification is given as Image size/ Object size

For calculations of magnification, values of image size and object size are taken with proper sign convention. 
For an inverted image, the image size is taken as negative and hence magnification is negative.

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

An astronomical telescope has an objective of focal length $200 \,cm$ and an eye piece of focal length $4\,cm$ The telescope is focused to see an object $10\, km$ from the objective,.The final image is formed at infinity. The length of the tube and angular magnification produced by it is

  1. $204\, cm, -50$
  2. $200\, cm, -50$
  3. $204\, cm, -100$
  4. $200\, cm, -100$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For an astronomical telescope, the length of the tube is fo + fe = 200 + 4 = 204 cm. The angular magnification for a distant object is -fo/fe = -200/4 = -50.

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

Magnification m = _____

  1. v/u

  2. u/v
  3. $h _0/h _i$
  4. $h _i/h _0$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

The magnification equation relates the ratio of the image distance(v) and object distance(u) and also to the ratio of the image height ($h _i$) and object height ($h _o$).

i.e., $m=\dfrac vu=\dfrac {h _i}{h _o}$

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

An astronomical telescope has focal lengths $100$ & $10$cm of objective and eyepiece lens respectively when final image is formed at least distance of distinct vision,magnification power of telescope will be,

  1. -15

  2. -14

  3. -17

  4. -19
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given focal length of eye piece${f} _{e}=10cm\$

focal length of objective${f} _{o}=100 cm\$
Also we know least distance $D=25 cm\$ 
Magnifying power $M=\dfrac{-{f} _{0}}{{f} _{e}}(1+\dfrac{{f} _{e}}{D})\$
$M=-\dfrac{100}{10}(1+\dfrac{10}{25})\$
$M=-14$

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

Magnification for erect and invented image is

  1. $+ve$ and $-ve$ respectively
  2. $-ve$ and $+ve$ respectively
  3. $+ve$
  4. $-ve$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Magnification is the ratio of height of image and object .

And by convention , height of image formed below principal axis is taken negative and above is taken positive.

Hence, for erect image, $m=+ve$ and for inverted image $m=-ve$.

Answer-(A).

Multiple choice zoology our environment - our concern biomagnification the effects of human activities on ecosystems bioaccumulation and biomagnification

Which is the correct pair of biological magnification?

  1. Water - $0.0003$ ppm
  2. Small fishes - $0.2$ ppm
  3. Big fishes - $2$ ppm
  4. Zoo plaktery - $0.4$ ppm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Biological magnification is also called as biomagnification and bioamplification. It occurs due to the increasing concentration of a toxic substance in the tissues of tolerant organisms that successively move up to higher levels in a food chain. The order of food chain for the given example is water followed by zoo plaktery followed by small fishes and lastly big fishes. It usually increase 2 to 10 folds in a food chain. Hence, the biomagnification of water, that is 0.0003 ppm is too less. Also, biomagnification of small fishes (0.2 ppm) should be ideally more than zoo plaktery (0.4 ppm). Hence, options A, B and D are incorrect. Big fishes having hightest biomagnification, that is 2 ppm.

Thus, the correct answer is 'Big fishes - 2 ppm.'

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

When a telescope is adjusted for normal vision, the distance of the objective from the eye-piece is found to be 80 cm. The magnifying power of the telescope is 19. What are the focal lengths of the lenses ?

  1. 61 cm, 19 cm

  2. 76 cm, 4 cm

  3. 40 cm, 40 cm

  4. 50 cm, 30 cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a telescope, L = fo + fe = 80 and M = fo/fe = 19. So fo = 19fe. 19fe + fe = 80, 20fe = 80, fe = 4. fo = 76.