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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

In an astronomical telescope, the intermediate image is

  1. virtual, erect and magnified

  2. real, erect and magnified

  3. real, inverted and reduced

  4. virual, inverted and reduced
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Astronomical telescope in which the image is inverted$,$ is n of the two principle types of the telescopes$.$ Its primary function is to enlarge the retinal image of a distant object$.$

hence,
option $(C)$ is correct answer.

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

The aperture of the largest telescope in the world is $5 m,$ if the separation between the Moon and the Earth is $4 \times 10^5 km$ and the wavelength of the visible light is $5000 \overset {o}{A}$ then the minimum separation between the objects on the surface of the Moon which can be just resolve is approximately

  1. $1 m$

  2. $10 m$

  3. $50 m$

  4. $200 m$
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

If an astronomical telescope has objective and eye-pieces of focal length 200 cm and 4 cm respectively,then the magnifying power of the telescope for the normal vision is:

  1. 42

  2. 50

  3. 58

  4. 204
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

 

It is given that,

Focal length of eye- piece fe = 4 cm

Focal length of object is fo = 200 cm

Least distance of distinct vision is d = 25 cm

So, magnifying power of microscope is

$ M=\dfrac{-{{f} _{0}}}{{{f} _{e}}}\left( 1+\dfrac{{{f} _{e}}}{d} \right) $

$ M=\dfrac{-200}{4}\left( 1+\dfrac{4}{25} \right) $

$ =-58\,cm $

 

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.