Tag: physics

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Multiple choice physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

For an electron microscope, which of the following is false ?

  1. It uses magnetic lens to converge electron beam.

  2. Its resolving power is directly proportional to accelerating potential of electron.

  3. Its resolving power is inversely proportional to wavelength of electrons

  4. Magnification attained with the help of of it of the order of $10^6$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For an electron microscope: (A) is correct - magnetic lenses are used to focus the electron beam. (C) is correct - resolving power is inversely proportional to electron wavelength (R ∝ 1/λ). (D) is correct - electron microscopes can achieve magnifications of ~10⁶. (B) is FALSE - resolving power is NOT directly proportional to accelerating potential; it increases with potential because higher potential reduces electron wavelength (λ = h/√(2meV)), but this is an inverse relationship, not direct proportionality.

Multiple choice physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

A telescope has an objective of focal length $50 cm$ and an eye-piece of focal length $5 m$ THe least distance of distinct vision is $25 cm$. The telescope is focused for distinct vission a scale $200 cm$ away from the objective.The separation between the two lenses is nearly 

  1. $71 cm$

  2. $61 cm$

  3. $81 cm$

  4. $51 cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics observing space: telescopes optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

Match the items given in A, B, C and D against their corresponding description given in 1, 2, 3, 4 and 5.
A. Astronomical telescope
B. Galilean Telescope
C. Simple Microscope
D. Compound Microscope

1. Contains one convex lens
2. Contains one concave lens
3. Contains a convex lens of large focal length in the objective and a convex lens of small focal length in the eye-piece
4. Contains a convex lens of small focal length in the objective and a convex lens of large focal length in the eye-piece
5. Contains a convex lens of large focal length in the objective and a concave lens of small focal length in the eye-piece. 

  1. $2\, \quad\, 3\, \quad\, 1\, \quad\, 5$

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

  3. $2\, \quad\, 5\, \quad\, 3\, \quad\, 4$

  4. $3\, \quad\, 5\, \quad\, 1\, \quad\, 4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A). Astronomical telescope - Contains a convex lens of large focal length in the objective and a convex lens of small focal length in the eye-piece

B.) Galilean Telescope - Contains a convex lens of large focal length in the objective and a concave lens of small focal length in the eye-piece
C). Simple Microscope - Contains one convex lens
D.) Compound Microscope - Contains a convex lens of small focal length in the objective and a convex lens of large focal length in the eye-piece
Thus,
A  -  3 , B  -  5 , C  -  1 and D  -  4 , hence Option D is correct.

Multiple choice physics observing space: telescopes optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

The sum of the focal lengths of the objective and an eyepiece, In case of an astronomical telescope is equal to : ( final image is at $\infty$)

  1. The length of the telescope

  2. Half the length of the telescope

  3. Double the length of the telescope

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Answer is A.

The astronomical telescope makes use of two positive lenses: the objective, which forms the image of a distant object at its focal length, and the eyepiece, which acts as a simple magnifier with which to view the image formed by the objective. Its length is equal to the sum of the focal lengths of the objective and eyepiece, and its angular magnification is -fo/fe , giving an inverted image.
Hence, the sum of the focal lengths of the objective and an eyepiece, In case of an astronomical telescope is equal to the length of the telescope.

Multiple choice physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

The focal lengths of the objective and the eyepiece of an astronomical telescope are 20 cm and 5 cm respectively. If the final image is formed at a distance of 30 cm from the eyepiece, find the separation between the lenses required for distinct vision

  1. 32.4 cm

  2. 42.3 cm

  3. 24.3 cm

  4. 30.24 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$f _{0}=20cm$


$f _{e}=5cm$

$V _{e}=30cm = D$

$L=?$

$L _{D}=f _{0}+ \dfrac{Df _e}{D+f _e}$ $=20+\dfrac{5\times 30}{35}=24.3 cm$

Multiple choice physics option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

The focal length of the objective of an astronomical telescope is 1 m and it is in normal adjustment.Initially the telescope is focussed to a heavenly body. If the same telescope is to be focussed to an object at a distance of 21 m from the objective,then identify the correct choice

  1. eye piece should be displaced by 2 cm away from the objective

  2. eye piece should be displaced by 2 cm towards the objective

  3. eye piece should be displaced by 5 cm towards the objective

  4. eye piece should be displaced by 5 cm away from the objective
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know, $ \dfrac {1}{f} = \dfrac {1}{v} - \dfrac {1} {u} $


We seeing heavenly body,  $u =  \infty$
$v = f = 1 m$

When seeing 21 m far
$u = - 21 m $
$f = 1 m$
$v = \dfrac {f u} {(f+u)} = 21 / 20 = 1.05 m$

So, eye piece need to move $1.05-1 = 0.05 m$ further away from objective

Answer. D) eye piece should be displaced by $5 cm$ away from the objective