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Questions Related to lenses

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

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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

Show answer
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$^{\circ}$

Show answer
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.

The difference between angle of minimum deviation for violet and red rays in the spectrum of white light from a prism is $2^0$. If the angle of minimum deviation of the mean ray is $48^0$, the dispersive power of the material of the prism is

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

  1. $24^0$

  2. $48^0$

  3. 0.0416

  4. 0.0832

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Correct Option: D

The focal length of a thin convex lens for red and blue rays are $100\ cm$ and $96.8\ cm$ respectively. The dispersive power of the material of the lens is

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

  1. 0.0325

  2. 0.825

  3. 0.968

  4. 0.98

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Correct Option: A

If a glass prism is dipped in water, its dispersive power

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

  1. increases

  2. decreases

  3. does not change

  4. may increase or decrease depending on whether the angle of the prism is less than or greater than $60^o$

Show answer
Correct Option: A
Explanation:

Dispersive power of a prism $=\dfrac{\mu _{v}-\mu _{r}}{\mu-1}$


Here the refractive indices are with respect to the medium in which prism is kept.

Hence when in water, dispersive power $=\dfrac{\dfrac{\mu _{v}}{\mu _{water}}-\dfrac{\mu _{r}}{\mu _{water}}}{\dfrac{\mu}{\mu _{water}}-1}$
$=\dfrac{\mu _{v}-\mu _{r}}{\mu-\mu _{water}}$

Hence dispersive power increases when prism is dipped in water.