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

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.

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

The diameter of the lens of a telescope is 1.22 m., the wavelength of light is $5000{A^0}$ the resolution power of the telescope is 

  1. $2 \times {10^5}$
  2. $2 \times {10^6}$
  3. $2 \times {10^2}$
  4. $2 \times {10^4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Resolving power is defined as D / (1.22 * lambda). With D = 1.22 m and lambda = 5e-7 m, RP = 1.22 / (1.22 * 5e-7) = 1 / 5e-7 = 2e6.

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

If an object subtend angle of $2^0$ at eye when seen through telescope having objective and eyepiece of focal length $f _0=60cm$ and $f _e=5cm$ respectively than angle subtend by image at eye piece will be

  1. $16^0$
  2. $50^0$
  3. $24^0$
  4. $10^0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The magnification M = f_o / f_e = 60 / 5 = 12. The angle subtended by the image is beta = M * alpha = 12 * 2 degrees = 24 degrees.

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

An astronomical telescope has an eye piece of focal length $5\ cm$. If magnification produced is 14 in normal adjustment, then calculate the length of the telescope.

  1. $75\ cm$
  2. $9\ cm$
  3. $50\ cm$
  4. $55\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In normal adjustment, M = f_o / f_e = 14. Given f_e = 5 cm, f_o = 14 * 5 = 70 cm. The length L = f_o + f_e = 70 + 5 = 75 cm.

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

A narrow vertical slit of width 2 mm is placed in front of a telescope.This setup is used to observe a car with its head light 1.2 m apart. The diameter of the objective of the telescope is 2 cm and the wavelength of light from the headlights is $ 5000 \mathring { A }  $. The distance of the car when the two headlights of the car are just resolved is:-

  1. 1.5 Km

  2. 2.6 Km

  3. 4.8 Km

  4. 3.93 Km
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

The ratio of resolving power of telescope, when lights of wavelength $4400\overset{o}{A}$ and $5500\overset{o}{A}$ are used, is _________.

  1. $16:25$
  2. $4:5$
  3. $9:1$
  4. $5:4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Resolving power $\infty \dfrac{1}{\lambda}$
$\dfrac{(R.P.) _1}{(R.P.) _2}=\dfrac{\lambda _2}{\lambda _1}=\dfrac{5500}{4400}=\dfrac{5}{4}$.

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

The diameter of the moon is $3.5\times 10^{3} km$ and its distance from the earth is $3.8\times 10^{5} km$. The diameter of the image of the moon seen by a telescope having focal length of the objective and eye-piece as $400\ cm$ and $10\ cm$ respectively will be

  1. $11^{\circ}$
  2. $21^{\circ}$
  3. $31^{\circ}$
  4. $41^{\circ}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Angular size of moon alpha = 3.5e3 / 3.8e5 = 0.0092 radians. Magnification M = f_o / f_e = 400 / 10 = 40. Image angle beta = 40 * 0.0092 = 0.368 radians, which is approximately 21 degrees.

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

The magnification produced by an Astronomical telescope in normal

  1. $f _{0}+f _{e}$
  2. $f _{0}\times f _{e}$
  3. $\dfrac{f _{ 0} }{ f _{e}}$
  4. $\dfrac{f _{e} }{ f _{0}}$
Reveal answer Fill a bubble to check yourself
C 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 the focal length of the eyepiece.

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

A telescope has a objective of focal length $60\, cm$ and an eye-piece of focal length $5\, cm$. The least distance of distinct vision is $25\, cm$. The telescope is focussed for distinct vision on a scale of $300\, cm$ away. The separation between the objective and the eye-piece is

  1. $71\, cm$
  2. $67\, cm$
  3. $83\, cm$
  4. $79\, cm$
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

The length of an astronomical telescope for normal vision (relaxed eye) will be:

  1. $f _0 - f _e$
  2. $f _0 / f _e$
  3. $f _0 \times f _e$
  4. $f _0 + f _e$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a relaxed eye, the image is formed at infinity, which corresponds to the normal adjustment of the telescope where the length is the sum of the focal lengths of the objective and eyepiece.

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

The focal length of the objective of a terrestrial telescope is $80cm$ and it is adjusted for parallel rays, then its power is $20$. If the focal length of erecting lens is $20cm$, then full length of the telescope will be

  1. $164cm$
  2. $124cm$
  3. $100cm$
  4. $84cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Magnification for parallel rays
$m=\cfrac { { f } _{ o } }{ { f } _{ e } } $
$\Rightarrow 20=\cfrac { 80 }{ { f } _{ e } } $
or ${ f } _{ e }=4cm$
If the focal length of erecting lens is $20cm$ then the length of the telescope
${ L } _{ \infty  }={ f } _{ o }+4f+{ f } _{ e }\quad $
[where $f$ is the focal length of erecting lens]
$=80+4\times 20+4=164cm$

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

An astronomical refractive telescope has an objective of focal length 20 m and an eyepiece of focal length 2 cm. then

  1. the magnification is 1000

  2. the length of the telescope tube is 20.02 m

  3. the image formed is inverted

  4. all of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Here, $f _0 = 20 \,m \,\, and \,\, f _e = 2 \, cm = 0.02 \, m$ In normal adjustment, Length of telescope tube, $L = f _0 + f _e = 20 +0.02 = 20.02m$
and magnification, $m = \dfrac{f _0}{f _e} = \dfrac{20}{0.02} = 1000$
The image formed is inverted with respect to the object.