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Questions Related to physics

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

Large aperture objective is used in telescope. Which of the following is not a function of the objective?

  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
B Correct answer
Explanation

Large aperture objective is used in telescope as it can help in increasing the brightness of image, increasing field of view and increasing intensity by gathering more light.

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.