Tag: observing space: telescopes

Questions Related to observing space: telescopes

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?

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

  1. $25 cm$

  2. $75 cm$

  3. $50 cm$

  4. $100 cm$

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Correct Option: B
Explanation:

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

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

Why is it advised to use telescope in a clear sky?

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

  1. Less interference between telescope and object.

  2. To get better quality of image

  3. To reduce the lateral shift in images caused by the environment

  4. All of the above

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Correct Option: D
Explanation:

In a clear sky, images formed by telescope are much sharper because of the following reasons:

1. Atmospheric interference is reduced significantly reduced and only the required object is projected.
2. Since rays don't change much speed on a clear day, image is distortionless and quality is improved.
3. On a clear sky, lateral displacement of object is either negligible or don't change much in small time intervals and this helps in studying the bodies well.

A telescope:

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

  1. converges light

  2. diverges light

  3. reflects light

  4. refracts light

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Correct Option: B
Explanation:

Telescope comprises of two convex lenses. 

Lens farther from eye behaves as converging lens and forms image at focal point. 
Lens nearer to eye behaves as diverging lens and forms a virtual image.
Hence, overall effect is diverging.

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.

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

  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.

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Correct Option: A,B,D
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.

Telescopes make the far objects appear:

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

  1. Farther

  2. Nearer

  3. Highly magnified

  4. Disappear

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Correct Option: B

The anguThe angular resolution of 10cm diameter telescope at a wavelength of $5000nm$ is of order of the

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

  1. $ 10^{4}$ rad

  2. $ 10^{-4}$ rad

  3. $ 10^{-6}$ rad

  4. $ 10^{6}$ rad

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Correct Option: C
Explanation:

Given that the Diameter of telescope is $D=10cm=0.1m$ and the wavelength $\lambda =5000nm$

Angular resolution $d\theta =\frac{1.22 \lambda}{D},$
$\Rightarrow d\theta = \frac{1.22 \times 5000 \times 10^{-9}}{0.1}=0.61 \times 10^{-6},$
Therefore it is in the order of $10^{-6},$
So the correct option is $C.$

The optical length of an astronomical telescope with magnifying power of 10, for normal vision is 44cm. What is focal length of the objective?

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

  1. 40cm

  2. 22cm

  3. 10cm

  4. 4cm

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Correct Option: A
Explanation:
Length of telescope $L = f _o + f _e$
$\therefore$ $f _o + f _e  = 44$
We get $f _e = 44 - f _o$
Magnification of astronomical telescope for normal vision $|M| = \dfrac{f _o}{f _e}$
OR $10 = \dfrac{f _o}{44 - f _o}$
OR $440 - 10 f _o = f _o$

$\implies$ $f _o = \dfrac{440}{11} =40$ cm

The final image formed by an astronomical telescope is:

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

  1. real and erect.

  2. virtual and erect.

  3. real and inverted.

  4. virtual and inverted.

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Correct Option: D
Explanation:

The objective lens produces a real, inverted image and the eyepiece acts as a simple magnifier and does not re-invert and produces a virtual image. So overall the image is inverted and virtual.

What is the magnifying power of a telescope whose objective and eye - piece have focal length 180 cm and 3 cm respectively?

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

  1. 50

  2. 40

  3. 30

  4. 60

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Correct Option: D
Explanation:

Here $F=180 \ cm$ and $f=30 \ cm$
$\therefore $ Magnifying power M
$=\dfrac {F}{f}=\dfrac {180}{3}=60$

The number of wavelengths in the visible spectrum is :

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

  1. $4000$

  2. $6000$

  3. $2000$

  4. $inifinite$

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Correct Option: D
Explanation:
  • The visible light spectrum is the section of the electromagnetic radiation spectrum that is visible to the human eye. 
  • It ranges in wavelength from approximately 400 nanometers ($4 \times  10 ^{-7}$ m, which is violet) to 700 nm ($7 \times 10^{-7}$ m, which is red).
  • Hence wavelengths are infinite
  • Option D is the right answer