Tag: ray optics and optical instruments

Questions Related to ray optics and optical instruments

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,


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

$= \dfrac {1}{20} + \dfrac {1}{30}$

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

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

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. 1.5D

  2. -1.5D

  3. 6.5D

  4. -2.5D

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Correct Option: B
Explanation:

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


$\dfrac{1}{f}=\dfrac{1}{40}+\dfrac{1}{-25}$

$f = -66.67 cm$

$P=\dfrac{100}{f}\ D$ $=-1.5 D$

A biconvex lens of focal length $f$ is cut into two plano convex lenses. If these two plano convex lenses are joined such that their convex surfaces are in contact, then the focal length of the combination is :

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

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

  2. $f/4$

  3. $2 f$

  4. $\text{infinite}$

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Correct Option: B
Explanation:

$f _{1}=f/2  $       $f _{2}=f/2$

$\dfrac{1}{f^{'}}=\dfrac{1}{f _{1}}+\dfrac{1}{f _{2}}$


$\dfrac{1}{f^{'}}=\dfrac{2}{f}+\dfrac{2}{f}$

$f'  = f/4$

Two identical plano convex lenses are arranged in three different combinations as:
I)  the plane surfaces touching each other
II) the curved surfaces touching each other
III) the plane surface of one lens touching the curved surface of the other.  
Then the focal lengths of the combinations are in the ratio:

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

  1. 1 : 2 : 3

  2. 1 : 1 : 1

  3. 3 : 2 : 1

  4. 2 : 2 : 1

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Correct Option: B
Explanation:

The focal lengths of the combination of two lenses kept in any fashion is irrespective of their arrangement. 

Two plano convex lenses of same material and of focal lengths 20 cm & 5 cm should be arranged such that the combination is free from chromatic aberration. Then effective focal length of the combination will be:

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

  1. 20 cm

  2. 4 cm

  3. 12.5 cm

  4. 8 cm

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Correct Option: C
Explanation:

The effective focal length of the lens is, 


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

Substituting the respective values, the effective focal length obtained is, $f = 12.5$