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 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 $4f$.

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

In an astronomical telescope, the distance between the objective and the eyepiece is 36 cm and the final image is formed at infinity. The focal length $f _0$ of the objective and the focal length $f _e$ of the eyepiece are

  1. $f _0=45 cm$ and $f _e=-9 cm$
  2. $f _0=50 cm$ and $f _e=22 cm$
  3. $f _0=65 cm$ and $f _e=7 cm$
  4. $f _0=30 cm$ and $f _e=6 cm$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

The focal lengths be $f _{o}$ and $f _{e}$ 


then $f _{o}+f _{e}=36$ for the image to be at infinity 

option $A$ is correct as $45-9=36$

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

State whether true or false :

Both the telescope and compound microscope make use of two lenses-one concave and one convex.

  1. True

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

Both the telescope and compound microscope make use of two lenses–one concave and one convex. [a] false

Both the lenses are convex lenses

The main difference between telescope and microscope is that microscopes are used to magnify small objects that are at a short distance from the viewer whereas telescopes are used to magnify large objects that are at a large distance from the viewer. In refracting telescopes, there are typically two convex lenses. One lens acts as the objective lens: this lens gathers light from faraway objects and forms a real, inverted image of the object at its focal point. A second lens, called the eyepiece

Microscopes are used to look at magnified images of small objects. A simple microscope consists of a single convex lens. The lens is held close to the object so that the object is between the lens and its focal point.

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

The focal length of eye lens and object lens of a telescope is 4 mm and 4 cm respectively. If final image of an far object is at $\displaystyle \infty $. Then the magnifying power and length of the tube are:

  1. 10, 4.4 cm

  2. 4, 44 cm

  3. 44, 10 cm

  4. 10, 44 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Magnification is the amount that a telescope enlarges its subject. Its equal to the telescopes focal length divided by the eyepieces focal length. As a rule of thumb, a telescopes maximum useful magnification is 50 times its aperture in inches (or twice its aperture in millimeters). 
That is, $M = fo / fe$
In this case, the focal length of eye lens and object lens of a telescope is 4 mm = 0.4 cm and 4 cm respectively.
So, Magnification $M = fo / fe = 4 / 0.4 = 10.$
Focal length of the eyepiece is the distance from the center of the eyepiece lens to the point at which light passing through the lens is brought to a focus.
Focal length of the objective is the distance from the center of the objective lens (or mirror) to the point at which incoming light is brought to a focus.
The length of the tube is given as sum of the focal lengths of the eye lens and the object lens.
So, Length of the tube $ = fo + fe = 4 + 0.4 = 4.4 cm.$
Hence, the magnifying power and length of the tube are: 10, 4.4 cm.

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

An astronomical telescope has an eyepiece of focal length $5 cm$. If magnification produced is $14$ in normal adjustment, then what is the length of telescope?

  1. $25 cm$
  2. $75 cm$
  3. $50 cm$
  4. $100 cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle m = \frac {f _o}{f _e}$ or $\displaystyle 14 = \frac {f _o}{5} \Rightarrow f _o = 70 cm$

$\displaystyle \therefore L=f _o + f _e = 5 + 70 = 75 cm$

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

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.

  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

Length of astronomical reflecting= distance between the objective and the eye - piece =L, 

$f _o=16m$ and $f _e=2cm=0.02m$
$L=f _o+f _e=16+0.02m=16.02m$
Angular magnification $m=\dfrac{f _o}{f _e}=\dfrac{16}{0.02}=800$. 
In astronomical reflecting telescope, the image formed is inverted.
The objective is always larger than eye - piece.