Tag: ray optics and optical instruments

Questions Related to ray optics and optical instruments

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. $-30cm$

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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. $+6.67D$

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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$

Two lenses of power $-15D$ and $-5D$ are in contact will each other. The focal length of the combination:

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

  1. $-20\ cm$

  2. $-10\ cm$

  3. $+20\ cm$

  4. $+10\ cm$

Show answer
Correct Option: A
Explanation:

Given that,

lenses power 

$P _{1}=-15\ D$

$P _{2}=-5\ D$

We know that,

$P=\dfrac{1}{f}$

Now,

  $ P={{P} _{1}}+{{P} _{2}} $

 $ P=-15-5 $

 $ P=-20 $

Now, the focal length is

  $ f=\dfrac{1}{P} $

 $ f=\dfrac{1}{-10} $

 $ f=0.02\,m $

 $ f=-20\,cm $

Hence the focal length is -$20\ cm$

There are two thin symmetrical lenses, one is converging with a refractive index  $2$ and the othe other is diverging with a refractive index $1.5$. Both lenses have same radius curvature of $10 cm$. The lenses were put together and submerged in water. What is the focal length of the system of water .The refractive index of water is $\cfrac{4}{3}$

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

  1. $40 cm$

  2. $\cfrac{40}{3} cm$

  3. $\cfrac{20}{3} cm$

  4. $-\cfrac{40}{3} cm$

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Correct Option: D

A convex lens of focal length 10 cm is in contact with a concave lens. The focal length of the combination is numerically equal to that of the concave lens. The focal length of the concave lens is :

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

  1. 10 cm

  2. 15 cm

  3. 5 cm

  4. 20 cm

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Correct Option: C

A diverging lens of focal length $-10cm$ is moving towards right with a velocity $5m/s$. An object, placed on principal axis is moving towards left with a velocity $3m/s$. The velocity of image at the instant when the lateral magnification produced is $1/2$ is: (All velocities are with respect to ground)

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

  1. $3m/s$ towards rigtht

  2. $3m/s$ towards left

  3. $7m/s$ towards rigtht

  4. $7m/s$ towards left

Show answer
Correct Option: A