Tag: option c: imaging

Questions Related to option c: imaging

The focal lengths of the objective and eyepiece of a telescope are 60cm and 5cm respectively.The telescope is focused on an object 360cm from the objective and the final image is formed at a distance of 30cm from the eye of the observer. The length of the telescope is

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

  1. 66.3 cm

  2. 86.3 cm

  3. 76.3 cm

  4. 96.3 cm

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Correct Option: C
Explanation:

$f _o = 60cm$


$f _e = 5cm$

$r _e = 30cm$

$L = ?$

$u _o = 360cm$

$\dfrac {1}{v _o} + \dfrac {1}{360} = \dfrac {-1}{60}$

$\dfrac {1}{v _o} = \dfrac {-1}{60} + \dfrac {1}{360}$

A planet is observed by an astronomical reflecting telescope having an objective of focal length $16 m$ and an eye-piece of focal length $2 cm$. Then :

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

  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:

A telescope uses two co-axially placed convex lenses in such a way that the focus of objective lens is past the focus of the eye piece as evident here from the focal lengths of the objective lens and the eye piece. $ \therefore $ The objective lens is larger than eye piece.

Length of the tube is given as 
$ L = f _o + f _e = 16\ m + 2\ cm = 16.02\ m $

Angular magnification is given as:
$ m = \dfrac{f _o}{f _e} = \dfrac{1600}{2} = 800 $


Since, the first lens or the objective lens produces a real and inverted image of the object to be observed and this real image formed acts as an object for the eye piece convex lens and is between the focus and the optical center, so, an inverted, virtual and enlarged image of the object is formed. $ \therefore $ The final image is inverted. 

Hence, the correct answers are OPTIONS A,B and D.

An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of

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

  1. large focal length and small diameter

  2. large focal length and large diameter

  3. small focal length and large diameter

  4. small focal length and small diameter

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Correct Option: B

The minimum value of Compton wavelength shift is:

detection and recording of x-ray images option c: imaging physics

  1. $h/2 m _{0}c$

  2. $h/m _{0}c$

  3. $2h/m _{0}c$

  4. $zero$

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Correct Option: A
Explanation:

$\lambda -\lambda ^{1}  =  \dfrac{h}{m _e c}\ (1+cos  \theta )$

$\Delta \lambda ^{1}  =  \dfrac{h}{m _e c}\ (1+cos  \theta )$

$cf   \theta = 0^{0}$

then  $\Delta \lambda = 0$

Consider a metal used to produced some characteristic $X-$rays. Energy of $X-$ray are given by $E$ and wavelength as represented by $\lambda$. Then which of the following is true:

detection and recording of x-ray images option c: imaging physics

  1. $E(K _{\alpha}) > E({K} _{\beta}) > E(K _{\gamma})$

  2. $E(M _{\alpha}) > E(L _{\alpha}) > E(K _{\alpha})$

  3. $\lambda (K _{\alpha}) > \lambda (K _{\beta}) > \lambda (K _{\gamma})$

  4. $\lambda (M _{ \alpha })>\lambda (L _{ \alpha })>\lambda (K _{ \alpha })$

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Correct Option: B

The shortest wavelength of X-rays emitted from an X-ray tube depends on

detection and recording of x-ray images option c: imaging physics

  1. The current in tube

  2. The voltage applied to the tube

  3. The nature of the gas in tube

  4. The atomic number of the target material

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Correct Option: B

The intensity of X-rays of wavelength $0. \mathring{A}$ reduces to one fourth on passing through $3.5 \ mm$ thickness of a metal foil. The coefficient of absorption of metal will be:-

detection and recording of x-ray images option c: imaging physics

  1. $0.2 \ mm^{-1}$

  2. $0.4 \ mm^{-1}$

  3. $0.6 \ mm^{-1}$

  4. $0.8 \ mm^{-1}$

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Correct Option: A

In Compton effect, the quantity $\dfrac{h}{m _{e}c}$ is called

detection and recording of x-ray images option c: imaging physics

  1. Compton recovery wavelength

  2. Scattered wavelength of photon

  3. Compton wavelength of electron

  4. Compton wavelength of photon

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Correct Option: C
Explanation:

$\lambda -\lambda ^{1}=\dfrac{h}{m _e c}(1-cos  \theta )$

where $\dfrac{h}{m _e c}$ is called compton wavelength.

The compton wavelength shift depends on

detection and recording of x-ray images option c: imaging physics

  1. Wavelength of the incident photon

  2. Material of the scatterer

  3. Energy of the incident photon

  4. Scattering angle

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Correct Option: D
Explanation:

$\lambda -\lambda ^{1}=\dfrac{h}{m _e c}\ (1+cos  \theta )$
where $\theta$ is scattering angle.

So, the answer is option (D).

Given $h = 6.62 \times 10^{-34}$ Js, $m _e$ $= 9.1 \times 10^{-31}$ kg, $c = 3 \times 10^{8}$ m/s, the value of Compton wavelength is:

detection and recording of x-ray images option c: imaging physics

  1. 0.0121 $A^{0}$

  2. 0.0484 $A^{0}$

  3. 0.0242 $A^{0}$

  4. 0.0363 $A^{0}$

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Correct Option: C
Explanation:

Compton wavelength $= \ \dfrac{h}{m _{e}c}$

$= \ \dfrac{6.62\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^{8}}$

$= \ 0.242\times 10^{-11}m$

$= \ 0.0242\times 10^{-10}m$

$= \ 0.0242\ A^{\circ}$