Tag: option c: imaging

Questions Related to option c: imaging

Advantages of optical fibres over electrical wires is:

physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

  1. High band width and EM interference

  2. Low band width and EM interference

  3. High band width low transmission capacity and no EM interference

  4. High band width, high data transmission capacity and no EM interference

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Correct Option: D
Explanation:

Few advantages of optical fibres are that the number of signals carried by optical fibres is much more than that carried by the Cu wire or radio waves. Optical fibres are practically free from electromagnetic interference and problem of cross talks whereas ordinary cables and microwave links suffer a lot from it.

(A): Optical fibres are widely used to communication network.
(R) : Optical fibres are small in size, light weight, flexible and there is no scope for interference in them.

physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

  1. Both (A) and (R) are true and (R) is the correct explanation of (A)

  2. Both (A) and (R) are true but (R) is not the correct explanation of (A)

  3. (A) is true but (R) is false

  4. (A) is false but (R) is true

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Correct Option: A
Explanation:

Optical fibres are widely used to communication network because they are small in size, light weight, flexible and there is no scope for interference in them. They are easy to handle due to it's small size, light weight and flexibility. They can be placed wherever we need. Further in this case, due to the no scope for interference , information cannot be loss or damage.

In optical fiber, refractive index of inner part is $1.68$ and refractive index of outer part is $1.44$. The numerical aperture of the fibre is

physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

  1. 0.5653

  2. 0.6653

  3. 0.7653

  4. 0.8653

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Correct Option: D
Explanation:

$Numerical$ $aperture$ $of$ $fibre$ $=$ $\sqrt{\mu _1 ^2 - \mu _2 ^2}$ $= \sqrt{1.68^2 - 1.44^2}$

$= \sqrt{2.8224 - 2.0736}$ $= \sqrt{0.7488} = 0.8653$ 

The angles of minimum deviations are 53$^{\circ}$ and 51$^{\circ}$ for blue and red colors respectively produced in an equilateral glass prism. The dispersive power is :

perceive colours resolution of optical instruments lenses option c: imaging physics

  1. $\dfrac{2}{51}$

  2. $\dfrac{1}{26}$

  3. $\dfrac{1}{52}$

  4. $\dfrac{1}{51}$

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Correct Option: B
Explanation:

As $\delta _{Y}=\dfrac{\delta _{B}+\delta _{R}}{2}$


So $\delta _{Y}=\dfrac{53^{\circ }+51^{\circ }}{2}$ =52$^{\circ }$

$\therefore dispersive\ power\ =\dfrac{\delta _{B}-\delta _{R}}{\delta _{Y}}$

                                    $=\dfrac{53-51}{52}$

                                    $=\dfrac{2}{52}$

                                    $=\dfrac{1}{26}$

If the refractive indices of crown glass for red, yellow and violet colours are respectively ${ \mu  } _{ r },{ \mu  } _{ y },{ \mu  } _{ v }$, then the dispersive power of this glass would be

perceive colours resolution of optical instruments lenses option c: imaging physics

  1. $\cfrac { { \mu } _{ v }-{ \mu } _{ r } }{ { \mu } _{ y }-1 } $

  2. $\cfrac { { \mu } _{ v }-{ \mu } _{ y } }{ { \mu } _{ r }-1 } $

  3. $\cfrac { { \mu } _{ v }-{ \mu } _{ y } }{ { \mu } _{ y }-{ \mu } _{ r } } $

  4. $\cfrac { { \mu } _{ v }-{ \mu } _{ r } }{ { \mu } _{ y } } -1$

Show answer
Correct Option: A
Explanation:

Dispersive power of a glass is given by ratio of difference of reflective index of two extreme wavelength to the difference of intermediate wavelength to unity i.e.


$ { Dispersive\quad Power\quad =\quad \dfrac { { \mu  } _{ v }-{ \mu  } _{ r } }{ { \mu  } _{ y }-1 }  } $

The dispersion of a medium for wavelength $\lambda$ is D. Then the dispersion for the wavelength $2\lambda$ will be:

perceive colours resolution of optical instruments lenses option c: imaging physics

  1. $(D/8)$

  2. $(D/4)$

  3. $(D/2)$

  4. D

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Correct Option: A
Explanation:

As we know, Cauchy's Dispersion formula is :
$ \mu $= $A + \dfrac{B}{ \lambda^{^{2}}} $
And dispersion
D= - $ \dfrac {d\mu}{d\lambda}$
Therefore, from the above 2 equations:
D = $ -(-2\lambda ^{3})B $= $ \dfrac {2B}{\lambda^{3}}$
This implies that
D $ \alpha \dfrac{1}{\lambda^{3}}$
Hence,
$ \dfrac {{D}'}{D}$ = $( \dfrac{\lambda}{{\lambda}' })^{3}$
As $ {\lambda}' = 2\lambda$
Therefore,
$ {D}' =D/8$

In a prism, the refractive indices of different colours are
$\mu _{V} =$ 1.6;
$\mu _{R} =$ 1.52;
$\mu _{Y} =$ 1.56.
The dispersive power of the prism is :

perceive colours resolution of optical instruments lenses option c: imaging physics

  1. $\dfrac{1}{56}$

  2. $\dfrac{1}{8}$

  3. $\dfrac{1}{7}$

  4. Infinite

Show answer
Correct Option: C
Explanation:

Dispersive power of prism= $w=\dfrac{\mu _{V}-\mu _{R}}{\mu _{Y}-1}$


$w=\dfrac{1.6-1.52}{1.56-1}$

$w=\dfrac{1}{7}$

A flint glass prism is of refracting angle 5$^{\circ}$. Its refractive index for C line is 1.790 and for F line is 1.805. The angular dispersion of C and F lines is:

perceive colours resolution of optical instruments lenses option c: imaging physics

  1. 0.075$^{\circ}$

  2. 0.085$^{\circ}$

  3. 0.095$^{\circ}$

  4. 0.065$^{\circ}$

Show answer
Correct Option: A
Explanation:

Deviation for C line $= \text{(Refractive index for C line -1) x Refracting angle}$

                                 $= (1.79-1) \times 5$
                                 $=3.95^o$

Deviation for F line $= \text{(Refractive index for F line -1) x Refracting angle}$

                                 $= (1.805-1) \times 5$      
                                 $= 4.025^o$

So the angular dispersion of C and F line is the deviation difference between F and C line.
Angular dispersion $= 4.025 -3.95 = 0.075^o$

Dispersive power of the material of a prism is 0.0221. If the deviation produced by it for yellow colour is 38$^{\circ}$, then the angular dispersion between red and violet colours is :

perceive colours resolution of optical instruments lenses option c: imaging physics

  1. 0.65$^{\circ}$

  2. 0.84$^{\circ}$

  3. 0.48$^{\circ}$

  4. 1.26$^{\circ}$

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Correct Option: B
Explanation:

As Dispersive power $ = \dfrac{Angular  \  dispersion}{mean  \ deviation}$


We take the deviation for yellow color as the mean deviation

$\Longrightarrow$  $\omega=\dfrac{\delta _{V}-\delta _{R}}{\delta _{Y}}$


$\Longrightarrow$  $0.0221=\dfrac{\delta _{V}-\delta _{R}}{38^{\circ }}$

$\Longrightarrow$  $\delta _{\nu}-\delta _{R}=0.84^{\circ }$

The dispersive power of a medium is

perceive colours resolution of optical instruments lenses option c: imaging physics

  1. The greatest for red light

  2. the least for red light

  3. the least for yellow light

  4. the ate for at colours

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Correct Option: B
Explanation:

We know $ P \propto\frac {1}{f}$ focal length is maximum for red light.