Tag: space research and satellites

Quizzes (0) Questions (96)

Questions Related to space research and satellites

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