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

The angular resolution of a 10cm diameter telescope at a wavelength of $5000 A^0$ is of the order of -

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

Given parameters are,

Wavelength $\lambda = 5000\ A^{\circ}$

 = $5000 \times 10^{-10}$

Diameter of telescope, $D = 10 cm = 0.1 m$

Now, Angular resolution (d\theta) formula for telescope is,

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

Substituting the values, we get

$\Rightarrow d\theta = \dfrac{1.22 \times 5000 \times 10^{-10}}{0.1}= 6.1 \times 10^{-6}\\$

Clearly it is having significance order of $10^{-6}$.

Thus option B is correct.

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 magnifying power of $20$. To the observer, the top 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 distant objects appear 20 times closer (or 20 times larger in angular size).

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 $\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

Answer is A.

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

Which of the following is correct about astronomical telescope?

  1. It consists of two diverging lenses

  2. Its objective is a concave lens

  3. Its eyepiece is a convex lens with greater focal length than the objective

  4. The final image in this telescope is inverted
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since in astronomical telescopes we are not much bothered about the inverted carterer. So, we get final image in this telescope is inverted where in terrestrial, we get upright image.

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

In an astronomical telescope the focal lengths of objective and eyepiece should respectively be :

  1. large and small

  2. small and large

  3. equal

  4. too small are too large
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In an astronomical telescope, two convex lenses are used for different focal lengths. In which the lens with large focal length is an objective lens which is used to see large distance objects like stars or planets and the one with small focal is eyepiece. Hence correct option is A.

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

While viewing a distant object with telescope a housefly sits on objective lens. Then which of the following is the correct statement :

  1. housefly will be seen enlarged in image

  2. housefly will be seen reduced in image

  3. intensity of image will be decreased

  4. intensity of image will be increased
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When housefly sits on the objective lens, light intensity will fall at objective because housefly will stop the light to fall on the objective. So, intensity of the image is decreased. Hence correct option is C.

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

Large aperture objective is used in telescopes as it helps in

  1. increasing the brightness of image

  2. reducing image size

  3. increasing field of view

  4. increasing intensity by gathering more light
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

The larger the objective, the more light the telescope collects and increases the brightness of image .

The field of view of the telescope decreases as the aperture increases, but the resolving power increases.
The objective lens of a telescope forms an real image of the night sky, the size of that image is in proportion to the focal length of the objective lens. It increases with increase in size of objective lens.

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

Large astronomical telescopes always use as objective

  1. lens

  2. mirror

  3. combinations of lenses

  4. none of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Large astronomical telescopes always use combination of lens as objective because
(1)To minimize spherical aberration.
(2)To maximize the amount of light ray entering the telescope.
(3)To optimize the quality of image.
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.