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

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:

    • 6.5 D
    • 6.5 D
    • 6.67 D
    • 1.5 D
Reveal answer Fill a bubble to check yourself
D Correct answer
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.
Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

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

  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. $\displaystyle \frac {f _1 - f _2}{f _1f _2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

  1. $0.33m$
  2. $1.33m$
  3. $2.33 m$
  4. $3 m$
Reveal answer Fill a bubble to check yourself
A Correct answer
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, $f = \dfrac {1}{3} = 0.33 m$

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

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

  1. $\dfrac{f}{2}$
  2. $f$
  3. $\dfrac{3f}{2}$
  4. $2f$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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

  1. $12\ cm$
  2. $60\ cm$
  3. $20\ cm$
  4. $30\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equivalent focal length,


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

$= \dfrac {1}{20} + \dfrac {1}{30}$

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

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

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

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 :

  1. $\dfrac{f}{2}$
  2. $f/4$
  3. $2 f$
  4. $\text{infinite}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$f _{1}=f/2  $       $f _{2}=f/2$

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


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

$f'  = f/4$

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

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:

  1. 1 : 2 : 3

  2. 1 : 1 : 1

  3. 3 : 2 : 1

  4. 2 : 2 : 1
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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:

  1. 20 cm

  2. 4 cm

  3. 12.5 cm

  4. 8 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
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, $f = 12.5$