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

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?

    • 15cm
    • 20cm
    • 60cm
    • 3 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. 50 cm

  2. 60 cm

  3. 90 cm

  4. 30 cm
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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  $       $f _{2}=f/2$

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


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

$f'  = f/4$