Tag: physics

Questions Related to physics

Match List-I with List- II and select the answer using the codes given below the lists:

List-I List-II
A. Microscope 1. To see objects on the surface by an observer in a trench
B. Telescope 2. To see small objects
C. Periscope 3. To see distant objects
D. Camera 4. To take photographs of objects
physics observing space: telescopes space research and satellites space travel space exploration and forms of light

  1. $2\, \quad\, 1\, \quad\, 4\, \quad\, 3$

  2. $1\, \quad\, 2\, \quad\, 3\, \quad\, 4$

  3. $2\, \quad\, 3\, \quad\, 4\, \quad\, 1$

  4. $2\, \quad\, 3\, \quad\, 1\, \quad\, 4$

Show answer
Correct Option: D
Explanation:

Microscope - is used to see small object which cannot be seen by naked eye

telescope - is used to see object which are situated at a long distance eg. sun, moon , planet
periscope -The uses of a periscope include observation around barriers and at times when viewing a location directly would be dangerous. Periscopes are typically used in submarines to see above the water.
Camera - is used to capture photos of objects.

Thus matching will be
A - 2 , B - 3   C - 1  and  D - 4

 hence Option D is correct.

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 $\displaystyle \infty $. Then the magnifying power and length of the tube are:

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

  1. 10, 4.4 cm

  2. 4, 44 cm

  3. 44, 10 cm

  4. 10, 44 cm

Show answer
Correct Option: A
Explanation:

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.

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. $100 cm$

Show answer
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

Show answer
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

Show answer
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.

Show answer
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

Show answer
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

Show answer
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.

Show answer
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.