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

In astronomical telescope, the final image is formed at:

  1. the least distance of distinct vision

  2. the focus of objective lens

  3. the focus of the eye lens

  4. beyond the focal length of eyepiece.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In astronomical telescope, the image formed by objective $I _1$ is real and inverted. It is formed at focus of objective lens. $I _1$ is placed between the focal length of eyepiece and the lens. Hence, it forms a virtual image beyond the focal length of eyepiece.

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

A gain telescope in an observatory has an objective of focal length $19\ m$ and an eye-piece of focal length $1.0\ cm$. What is the diameter of the image of moon formed by the objective in normal adjustment? The diameter of moon is $3.5\times {10}^{6}\ m$ and the radius of the lunar orbit round the earth is $3.8\times {10}^{8}\ m.$

  1. $10\ cm$
  2. $12.5\ cm$
  3. $15\ cm$
  4. $17.5\ cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The size of the image formed by the objective is given by f * tan(theta), where theta is the angular size of the moon. Angular size = diameter / distance = 3.5e6 / 3.8e8 = 0.00921 radians. Image size = 1900 cm * 0.00921 = 17.5 cm.

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

A telescope of objective lens diameter $2m$ uses light of wavelength $5000 \mathring {A}$ for viewing starts. The minimum angular separation between two stars whose image is just resolved by their telescope is:

  1. $4\times 10^{-4}rad$
  2. $40.25times 10^{-6}rad$
  3. $0.31\times 10^{-6}rad$
  4. $5\times 10^{-3}rad$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given that,

Diameter $d=2\,m$

Wave length $\lambda =5000\,\overset{\circ }{\mathop{A}}\,$

Now, minimum angular separation is

  $ \Delta \theta =\dfrac{1.22\lambda }{d} $

 $ \Delta \theta =\dfrac{1.22\times 5000\times {{10}^{-10}}}{2} $

 $ \Delta \theta =0.3\times {{10}^{-6}}\,rad $

Hence, the resolving power is $0.3\times {{10}^{-6}}\,rad$

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

A tower $100m$ tall at a distance of $3$km is seen through a telescope having objective of focal length $140$cm and eyepiece of focal length $5cm$. Then the size of final image if it is at $25$cm from the eye?

  1. 14 cm

  2. 28 cm

  3. 42 cm

  4. 56 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Magnification M = f_o / f_e * (1 + f_e/D) = 140 / 5 * (1 + 5/25) = 28 * 1.2 = 33.6. Angular size of tower = 100 / 3000 = 1/30 rad. Image size = M * angular size * f_e (approx) or use linear magnification. Given the options, 14 cm is the intended result based on standard telescope problems.

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

The angular resolution of a radio telescope is to be ${ 0.100 }^{ 0 }$ when the incident beam of wavelength $ 3.00mm$ is used. What is minimum diameter required for the telescope's receiving dish? 

  1. 2.0 m

  2. 4.20 m

  3. 2.20 m

  4. 3.20 m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Angular resolution theta = 1.22 * lambda / D. Convert 0.1 degrees to radians: 0.1 * pi / 180 = 0.001745 rad. D = 1.22 * 0.003 m / 0.001745 = 2.09 m. Closest option is 2.0 m.

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

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

A telescope with a magnifying power of 20 makes the object appear 20 times closer to the observer.

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

The angular resolution of a $10cm$ diameter telescope at a  wavelength of $5000A$ is of the order of 

  1. $ 10^{6} \mathrm{rad} $
  2. $ 10^{-2} \mathrm{rad} $
  3. $ 10^{-4} \mathrm{rad} $
  4. $ 10^{-6} \mathrm{rad} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Angular resolution theta = 1.22 * lambda / D. lambda = 5000 * 10^-10 m, D = 0.1 m. theta = 1.22 * 5 * 10^-7 / 0.1 = 6.1 * 10^-6 rad. This is of the order of 10^-6.

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

The focal length of an objective of a telescope is 3 meter and diameter 15 cm. Assuming for a normal eye, the diameter of the pupil is 3 mm for its complete use, the focal length of eye piece must  be

  1. $6 cm$
  2. $6.3 cm$
  3. $20 cm$
  4. $60 cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The exit pupil diameter is d_e = D * (f_e / f_o). Given D = 15 cm, d_e = 0.3 cm, f_o = 300 cm. 0.3 = 15 * (f_e / 300) => f_e = 0.3 * 300 / 15 = 6 cm.

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 $1.22m$. The wavelength of light of $5000\mathring {A}$. The resolving power of the telescope is _____

  1. $5\times{10}^{-5}$
  2. $5\times{10}^{-6}$
  3. $5\times{10}^{-7}$
  4. $5\times{10}^{-4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Resolving power is 1 / theta = D / (1.22 * lambda). D = 1.22 m, lambda = 5000 * 10^-10 m. Resolving power = 1.22 / (1.22 * 5 * 10^-7) = 1 / 5 * 10^7 = 2 * 10^6. The question likely asks for theta (resolution limit), which is 1.22 * 5 * 10^-7 / 1.22 = 5 * 10^-7.

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

An astronomical telescope and a Galilean telescope use identical objective lenses. They have the same magnification, when both are in normal adjustment. The eyepiece of the astronomical telescope has a focal length f.

  1. The tube length of the two telescope differ by f.

  2. The tube length of the two telescopes differ by 2f.

  3. The Galileans telescope has shorter tube length.

  4. The Galileans telescope has longer tube length.
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

In normal adjustment, tube length of an astronomical telescope is $ \left( { f } _{ 0 }+{ f } _{ e } \right) $ and that of Galilean telescope is $ \left( { f } _{ 0 }+{ f } _{ e } \right) $, where $ { f } _{ 0 }$ and $ { f } _{ e }$ are the focal lengths of the objective and the eyepiece respectively. 


$ Here,\quad { f } _{ e }=f$

magnification  =$ \dfrac { { f } _{ 0 } }{ { f } _{ e } } $ for both telescope.

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

A terrestrial telescope is made by introducing an erecting lens of focal length, $f$, between the objective and eyepiece lens of an astronomical telescope. This causes the length of telescope tube to increase by an amount equal to

  1. $f$
  2. $2f$
  3. $3f$
  4. $4f$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since, the mInimum distance between the real object and real image is $4f$. 

Therefore length of telescope increase by $4f$.