Tag: ray optics and optical instruments

Quizzes (21) Questions (129)

Questions Related to ray optics and optical instruments

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.

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

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)

  1. $3m/s$ towards rigtht
  2. $3m/s$ towards left
  3. $7m/s$ towards rigtht
  4. $7m/s$ towards left
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the lens formula 1/v - 1/u = 1/f and magnification m = v/u = 1/2. So v = u/2. 2/u - 1/u = -1/10, so 1/u = -1/10, u = -10, v = -5. Velocity of image v_i = m^2 * v_o = (1/2)^2 * 3 = 0.75. Considering relative velocities, the result is 3m/s towards the right.