Tag: combination of lenses

Questions Related to combination of lenses

Two thin lenses of focal length $f _1$ and $f _2$ are in contact and coaxial. The power of the combination is

physics ray optics and optical instruments power of the lens thin lens combination of lenses

  1. $\sqrt{\frac{f _1}{f _2}}$

  2. $\sqrt{\frac{f _2}{f _1}}$

  3. $\frac{f _1+f _2}{f _1f _2}$

  4. $\frac{f _1-f _2}{f _1f _2}$

Show answer
Correct Option: C
Explanation:

For the lenses placed coaxially, 

We have the relationship $P=P _1+P _2+P _3+......$ 
So,for given case P=$P _1+P _2$ or $P $= $\dfrac{1} {f _1} $+ $\dfrac{1} {f _2} $

So,C is the correct answer. 

Two thin lenses of focal lengths 20 cm and -20 cm are placed in contact with each other. The combination has a focal length equal to

physics ray optics and optical instruments power of the lens thin lens combination of lenses

  1. Infinite

  2. $50 cm$

  3. $60 cm$

  4. $10 cm$

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Correct Option: A

A glass slab of thickness 3 cm and refractive index 3/2 is placed on ink mark on a picec of paper, For a person looking at the mark at a distance 2 cm above it, the distance of the mark will paper to be 

physics ray optics and optical instruments power of the lens thin lens combination of lenses

  1. 3 cm

  2. 4 cm

  3. 4.5 cm

  4. 5 cm

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Correct Option: B

Three lenses have a combined power of $2.7 D$. If the powers of two lenses are $2.5 D$ and $1.7 D$ respectively, find the focal length of the third lens.

physics ray optics and optical instruments power of the lens thin lens combination of lenses

  1. $-66.66 cm$

  2. $-6.666 cm$

  3. $-66.66 m$

  4. $-6.666 m$

Show answer
Correct Option: A
Explanation:

Total power, $P _{net} = P _1 + P _2 + P _3$


$P _3 = P _{net} - P _1 - P _2$

$P _3 = 2.7 - 2.5 - 1.7$

      $= - 1.5 = \dfrac{1}{f _3}$

${f _3} = - \dfrac{1}{1.5} = -0.666m$

       $= -66.66cm$

Abeam of a parallel rays is brought to a focus by convex lens. If a thin concave lens of equal focal length is joined to the convex lens, the focus will

physics ray optics and optical instruments power of the lens thin lens combination of lenses

  1. Be shifted to infinity

  2. Be shifted by a small distance

  3. Remain undisturbed

  4. None of the above

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Correct Option: A
Explanation:

Abeam of a parallel rays is brought to a focus by convex lens. Now, when thin concave lens of equal focal length is joined to first lens, then combined focal length be

$\dfrac 1F=\dfrac 1{F _1}+\dfrac 1{F _2}=\dfrac 1f-\dfrac 1f=0[\because F _1=f, F _2=-f]\\implies F=\infty$
Thus, the image can be focused on infinity or focus shifts to infinity.

A symmetric double convex lens is cut into two equal parts along a plane perpendicular to the principal axis. If the power of the original lens is 4D, the power of the two pieces is :

physics ray optics and optical instruments power of the lens thin lens combination of lenses

  1. 2D

  2. 3D

  3. 4D

  4. 5D

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Correct Option: A
Explanation:

$P _{original} = 4D$


$P = P _{1}+P _{2}$

$\because $ convex lens is cut into two equal  parts

So, $P _{1}=P _{2}=P$

$P _{original} =P+P$

$4D= 2P$

$P=2D$

The focal length of the combination of two convex lens in contact is $f$ and if they are separated by a distance, then focal length of the combination is ${f} _{1}$. The correct statement is

physics ray optics and optical instruments power of the lens thin lens combination of lenses

  1. $f> {f} _{1}$

  2. $f={f} _{1}$

  3. $f< {f} _{1}$

  4. $f{f} _{1}=1$

Show answer
Correct Option: C
Explanation:

$f$ will be less than $f _1$


$Explanation$ 

$\dfrac{1}{f}= \dfrac {1}{F _1}  + \dfrac {1}{F _2}$

$ \dfrac{1}{f _1}= \dfrac {1}{F _1} + \dfrac{1}{F _2} - \dfrac{d}{F _1F _2}$
where $d$ is the distance between lenses.

Option C is correct.

Two thin lens of focal lengths ${f} _{1}$ and ${f} _{2}$ are in contact. The focal length of this combination is

physics ray optics and optical instruments power of the lens thin lens combination of lenses

  1. $\cfrac { { f } _{ 1 }{ f } _{ 2 } }{ { f } _{ 1 }-{ f } _{ 2 } } $

  2. $\cfrac { { f } _{ 1 }{ f } _{ 2 } }{ { f } _{ 1 }+{ f } _{ 2 } } $

  3. $\cfrac {2 { f } _{ 1 }{ f } _{ 2 } }{ { f } _{ 1 }-{ f } _{ 2 } } $

  4. $\cfrac {2 { f } _{ 1 }{ f } _{ 2 } }{ { f } _{ 1 }+{ f } _{ 2 } } $

Show answer
Correct Option: B
Explanation:

If resulting focus is $f$ then $ \dfrac{1}{f} = \dfrac{1}{f _1} + \dfrac{1}{f _2} $


which lead us to $f= \dfrac{f _1 f _2}{f _1 +f _2}$ 
Option B is correct.

Two thin lenses, one of focal length $+60cm$ and the other of focal length $-20cm$ are put in contact. The combined focal length is

physics ray optics and optical instruments power of the lens thin lens combination of lenses

  1. $+15cm$

  2. $-15cm$

  3. $+30cm$

  4. $-30cm$

Show answer
Correct Option: D
Explanation:

If combined focal length is $f$  then $ (1/f)  = \dfrac{1}{60} +\dfrac{1}{-20}= (-1/30)$ so $f=-30cm$

Option D is correct.

A convex lens of focal length $40$ cm is in contact with a concave lens of focal length $25$ cm. The power of combination is

physics ray optics and optical instruments power of the lens thin lens combination of lenses

  1. $-1.5D$

  2. $-6.5D$

  3. $+6.5D$

  4. $+6.67D$

Show answer
Correct Option: A
Explanation:

Power  = $ \cfrac{1}{F} = \cfrac{1}{f _1} + \cfrac{1}{f _2}$

 = $ \cfrac {1}{+0.4m} + \cfrac{1}{-0.25m}$
$ \cfrac{1}{F} = \cfrac{-0.25+0.4}{0.4 \times (-0.25)}$
$ \therefore P = \cfrac{1}{F} = \cfrac {0.15}{-0.1} = -1.5D$