Tag: physics

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Questions Related to physics

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

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 

  1. 3 cm

  2. 4 cm

  3. 4.5 cm

  4. 5 cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The apparent shift due to a glass slab is given by t(1 - 1/n). Here, 3(1 - 1/(3/2)) = 3(1 - 2/3) = 3(1/3) = 1 cm. The mark appears 1 cm closer to the surface. Since the person is 2 cm above the slab (total distance 5 cm from the mark), the apparent distance is 5 - 1 = 4 cm.

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

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.

  1. $-66.66 cm$
  2. $-6.666 cm$
  3. $-66.66 m$
  4. $-6.666 m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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


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

$P _3 = 2.7 - 2.5 - 1.7$

      $= - 1.5 = \dfrac{1}{f _3}$

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

       $= -66.66cm$

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

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

  1. Be shifted to infinity

  2. Be shifted by a small distance

  3. Remain undisturbed

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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 :

  1. 2D

  2. 3D

  3. 4D

  4. 5D
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$P _{original} = 4D$


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

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

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

$P _{original} =P+P$

$4D= 2P$

$P=2D$

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

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

  1. $f> {f} _{1}$
  2. $f={f} _{1}$
  3. $f< {f} _{1}$
  4. $f{f} _{1}=1$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  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 } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. $-1.5D$
  2. $-6.5D$
  3. $+6.5D$
  4. $+6.67D$
Reveal answer Fill a bubble to check yourself
A Correct answer
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$

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

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

  1. $-20\ cm$
  2. $-10\ cm$
  3. $+20\ cm$
  4. $+10\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 -$20\ cm$

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

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 $\cfrac{4}{3}$

  1. $40 cm$
  2. $\cfrac{40}{3} cm$
  3. $\cfrac{20}{3} cm$
  4. $-\cfrac{40}{3} cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the lens maker's formula 1/f = (n-1)(1/R1 - 1/R2). For the converging lens in water: (2/(4/3) - 1)(1/10 - (-1/10)) = (0.5)(0.2) = 0.1. For the diverging lens in water: (1.5/(4/3) - 1)(-1/10 - 1/10) = (0.125)(-0.2) = -0.025. Total power = 0.1 - 0.025 = 0.075. f = 1/0.075 = 40/3. Wait, the sign convention results in -40/3.