Tag: ray optics and optical instruments

Questions Related to ray optics and optical instruments

Two lenses  of power  6 D and -2D are combined  to form a single lens.  The focal length of  this  lens will be 

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

  1. $\frac{3}{2}m$

  2. $\frac{1}{4}m$

  3. 4 m

  4. $\frac{1}{8}m$

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Correct Option: C

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

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

  1. -1.5 dioptres

  2. -6.5 dioptres

  3. +6.5 dioptres

  4. +6.67 dioptres

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Correct Option: A

A convex lens is in contact with a concave lens. The magnitude of the ratio of their focal lengths is 2/3. Their equivalent focal length is 30 cm. What are their individual focal lengths?

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

  1. -15, 10

  2. -10, 15

  3. 75, 50

  4. -75, 50

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Correct Option: A
Explanation:

Let focal length of convex lens be $f$


Then focal length of the concave lens $ =  - 1.5f$


Equivalent focal length $ =  \dfrac{- 1.5 f \times f }{ (-0.5f) }= 3f$
 
=>$ f = 10 cm$

So individual focal lengths are 10 cm and -15 cm

A convex lens of focal length $f _1$ and a concave lens of focal length $f _2$ are placed in contact. The focal length of the combination is

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

  1. $(f _1 + f _2)$

  2. $(f _1 - f _2)$

  3. $\displaystyle \frac{f _1 f _2}{f _2 - f _1}$

  4. $\displaystyle \frac{f _1 f _2}{f _2 + f _1}$

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Correct Option: C
Explanation:

$\displaystyle \frac{1}{f} = \frac{1}{f _1}+ \frac{1}{- f _2} = \frac{f _2 - f _1}{f _1 f _2} ;        f = \frac{f _1f _2}{f _2 - f _1}$

A lens of power +2 diopter is placed in contact with a lens of power -1 diopter. The combination will behave like

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

  1. a convergent lens of focal length 50 cm

  2. a divergent lens of focal length 100 cm

  3. a convergent lens of focal length 100 cm

  4. a convergent lens of focal length 200 cm

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Correct Option: C
Explanation:

$P = P _1 + P _2 = + 2 - 1 = +1$ diopter, lens behaves as convergent
$F = \displaystyle \frac{1}{P} = \frac{1}{1} = 1m = 100 cm$

The radius of curvature of the convex surface of a plano-convex lens is $10 cm$. What is the focal length of the plano-convex lens? (Here $\displaystyle \mu $ = $1.5$) 

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

  1. $10 cm$

  2. $20 cm$

  3. $15 cm$

  4. $5 cm$

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Correct Option: B
Explanation:

$\mu=1.5 , R _1=10cm , R _2=0$


As focal length is given by $\dfrac{1}{f}=(\mu-1)(\dfrac{1}{R _1}-\dfrac{1}{R _2})$  (lens maker formula)

$\dfrac{1}{f}=0.5\times (\dfrac{1}{10}-0)$

$f=20cm$

hence, the focal length of the plano-convex lens is $20cm$.

Two thin lenses of focal lengths 20 cm and 25 cm are placed in contact. The effective power of the combination is :

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

  1. $\displaystyle \frac{1}{9}$ diopters

  2. 45 diopters

  3. 6 diopters

  4. 9 diopters

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Correct Option: D
Explanation:

$f _1=20cm=0.2m ; f _2=25cm=0.25m$
as power of lens is given by,
$P=\frac{1}{f}$
$P _1=\frac{1}{f _1}=\frac{1}{0.2}=5D$,
$P _2=\frac{1}{f _2}=\frac{1}{0.25}=4D$
so,The effective power of the combination is $5D+4D=9D$.
hence,option D is correct.

The power of a diverging lens is 10D and that of a converging lens is 6D. When these two lenses are placed in contact with each other. The power of their combination will be

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

  1. +16D

  2. -16D

  3. -4D

  4. +4D

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Correct Option: C
Explanation:
The focal length of diverging lens is
${ F } _{ 1 }=\dfrac { 1 }{ P } =-\dfrac { 1 }{ 10 }$
The focal length of converging lens is
${ F } _{ 2 }=\dfrac { 1 }{ P } =+\dfrac { 1 }{ 6 }  $
The focal length of combined lens is
$\dfrac { 1 }{ f } =\dfrac { 1 }{ { f } _{ 1 } } +\dfrac { 1 }{ { f } _{ 2 } } $

$P=\dfrac { 1 }{ f } =-10+6=-4D$
The combined lens will act as diverging (concave ) lens.
Option C is correct.


The power of a combination of two lenses A and B is 5D.The focal length of A is 15cm. What is the focal length of B?

physics ray optics and optical instruments power of the lens thin lens combination of lenses
    • 15cm
    • 20cm
    • 60cm
    • 3 cm
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Correct Option: C
Explanation:

Answer is C.

Assuming the two lenses are in contact we have 1/f = 1/f1 + 1/f2 
where the total power 1/f = power = 5D 
We know, P = 1 / f (that is, 1D) where f = 1 m. The units of dioptre is 1/m .
Therefore, 100/15 + 1/f2 = 5 
1/f2 = -25/15 
f2 = -15/25 
= -0.6 = -60 cm.
Hence, the focal length of B is -60 cm.

What is the focal length of double convex lens for which the radius of curvature of each surface is 60 cm ( n= 1.5)

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

  1. 50 cm

  2. 60 cm

  3. 90 cm

  4. 30 cm

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Correct Option: B
Explanation:

$R _1$ = 60 cm, $R _2$ = -60 cm, n = 1.5
[$\because$ For a double convex lens, $R _2 =-1R _1$]
$\therefore \displaystyle \frac{1}{f} =(n-1) \left[ \frac{1}{R _1 -{R _2}}\right]$
$\Rightarrow \displaystyle \frac{1}{f} = (1.5-1) \left[ \frac{1}{60} +\frac{1}{60}\right]$
$\Rightarrow \displaystyle \frac{1}{f} = 0.5 \times \frac{2}{60} = 0.5 \times \frac{1}{30} =\frac{1}{60}$
$\Rightarrow $  f= 60 cm