Tag: power of the lens

Questions Related to power of the lens

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

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
    • 6.5 D
    • 6.5 D
    • 6.67 D
    • 1.5 D
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Correct Option: D
Explanation:
We know, the focal length of lenses in contact is given by
$ \dfrac{1}{F} = \dfrac{1}{f _1} + \dfrac{1}{f _2} +  ....  + \dfrac{1}{f _n}$

Here, 
For convex lens, $ f _1 = 40\ cm $
For concave lens, $ f _2 = -25\ cm $

$ \dfrac{1}{F} = \dfrac{1}{40} + \dfrac{1}{-25} =  \dfrac{1}{40} - \dfrac{1}{25} = \dfrac{25 - 40}{40 \times 25} = \dfrac{-15}{1000} $

$ F = \dfrac{-1000}{15}\ cm $

Power of the combination is given by

$ P = \dfrac{100}{F\ (in\ cm)} = \dfrac{-15 \times 100}{1000}  =\dfrac{-15}{10} = -1.5 D $

Thus, power of the combination is $ -1.5\ D $.

Hence, the correct answer is OPTION D.

Two lenses of power $-15\ D$ and $+5\ D$ are in contact with each other. The focal length of the combination is

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. $+\ 10\ cm$

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Correct Option: B
Explanation:

effective power is $-15D+5D=10D$


focal length is $\dfrac{1}{10}\times 100=10$

option $B$ is correct 

If two thin lenses of focal length $f _1$ and $f _2$ are kept in contact to each other, then the equivalent focal length of the combination

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

  1. $\displaystyle \frac {f _1f _2}{f _1 + f _2}$

  2. $\displaystyle \frac {f _1f _2}{f _1 - f _2}$

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

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

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Correct Option: A
Explanation:

Net Power of a combination of lens is sum of their individual power i.e.
$P _{net}=P _{1}+P _{2}$
$\dfrac{1}{f _{net}}=\dfrac{1}{f _{1}}+\dfrac{1}{f _{2}}$
$f _{net}=\dfrac{f _{1}+f _{2}}{f _{1}f _{2}}$
$f _{net}=\dfrac{f _{1}f _{2}}{f _{1}+f _{2}}$

Two thin lenses of power $2D$ and $1D$ are placed in contact. Find the focal length of the lens combination.

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

  1. $0.33m$

  2. $1.33m$

  3. $2.33 m$

  4. $3 m$

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Correct Option: A
Explanation:

Power of the lens combination, $P = P _{1} + P _{2}$
$= 2D + 1 D$
$= 3D$
$P = \dfrac {1}{f}$
$3 = \dfrac {1}{f}$
Therefore, focal length of lens combination, $f = \dfrac {1}{3} = 0.33 m$

The equiconvex lens has focal length $f$. If it is cut perpendicular to the principal axis passing through optical centre, then focal length of each half is

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

  1. $\dfrac{f}{2}$

  2. $f$

  3. $\dfrac{3f}{2}$

  4. $2f$

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Correct Option: D
Explanation:

When an equiconvex lens is cut parallel to principle axis focal length remains the same and when the lens is cut perpendicular to principle axis its focal length becomes twice the original.

The plano-convex lens of focal length $20\ cm$ and $30\ cm$ are placed together to form a double convex lens. The final focal length will be

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

  1. $12\ cm$

  2. $60\ cm$

  3. $20\ cm$

  4. $30\ cm$

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Correct Option: A
Explanation:

Equivalent focal length,


$\dfrac {1}{F} = \dfrac {1}{f _{1}} + \dfrac {1}{f _{2}}$

$= \dfrac {1}{20} + \dfrac {1}{30}$

$\therefore F = \dfrac {20\times 30}{20 + 30}$

$= \dfrac {600}{50} = 12\ cm$