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

The ratio of the size of the image to the size of the object is known as :

  1. the focal plane

  2. the transformation ratio

  3. the efficiency

  4. the magnification ratio
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

magnification ratio is given as:

 size or height of image/ size or height of object
substituted with proper sign convention.

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 wave optics huygens wave theory and wavefront wave propagation (huygens' construction) theories on light wave behaviour

 For what distance ray optics a good approximation when the aperture is 4 mm wide and the wavelength is 500nm?

  1. $32m$
  2. $69 m$
  3. $16 m$
  4. $8 m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The Good approximation distance till ray optics or valid is given by Fresnel distance which is given by
$z _f=\dfrac{a^2}{\lambda}$
$z _f$ is distance till which diffraction effects of light can be neglected.
$a=$ aperature$=4\times 10^{-3}m$
$\lambda =$wavelength$=500\times 10^{-9}$m
$z _f=\dfrac{(4\times 10^{-3})^2}{(500\times 10^{-9})m}$
$=\dfrac{16\times 10^{-6}}{500\times 10^{-9}}$
$=0.032\times 10^3$m
$\therefore z _f=32m$.
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 $(m) =$ ______

  1. $\dfrac {v}{u}$
  2. $\dfrac {u}{v}$
  3. $\dfrac {h _{o}}{h _{i}}$
  4. $\dfrac {h _{i}}{h _{o}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Magnification is ratio of the size of the image $h _i$ to the size of the object $h _o$.

$m=\dfrac{h _i}{h _o} = \dfrac{-v}{u}$ where $u$ is object distance and $v$ is image distance.

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

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. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$.........(1)


multiplyng by u in eq.(1)

$\dfrac{u}{f}=\dfrac{u}{v}+\dfrac{u}{u}$

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

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

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

as $m=\dfrac{v}{u}$

hence, $m=\dfrac{f}{u-f}$

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

The linear magnification for a spherical 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. $\dfrac {u}{u-f}$
  2. $\dfrac {u f}{u-f}$
  3. $\dfrac {f}{u+f}$

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

General equation for a spherical mirror says that:
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

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

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

$\dfrac{v}{u}=\dfrac{f}{u+f}=m$ (magnification)

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

A flim projector magnifies a flim of area $100 $ square centimeter on screen. If linear magnification is $4$ then area of magnified image on screen will be-

  1. $1600 sq. cm$
  2. $800 sq. cm$
  3. $400 sq. cm$
  4. $200 sq. cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As linear magnification, $M=4$

Hence, a real magnification ${ m } _{ r }={ m }^{ 2 }$
${ \left( 4 \right)  }^{ 2 }=16$
Surface area of film image on screen $=16\times 100=1600$ ${ cm }^{ 2 }$.

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.

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

In displacement method,magnification for two positions of the lens are 2 and 0.5 and the distance between the two position of the lens is 30 cm. if the focal length of lens is

  1. 15cm

  2. 20cm

  3. 25cm

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

In the displacement method, the focal length f is given by f = d(1-m^2) / (m1-m2) is incorrect; the correct formula is f = D*m / (1+m)^2 or using the displacement d and magnifications m1=2, m2=0.5. Since m1*m2 = 1, the object distance u and image distance v are swapped. The distance between positions is d = v-u = 30. With m=v/u=2, v=2u. Then 2u-u=30, so u=30, v=60. Focal length f = (u*v)/(u+v) = (30*60)/(90) = 20 cm.