Tag: combination of lenses

Questions Related to 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 

  1. 1.5D

  2. -1.5D

  3. 6.5D

  4. -2.5D


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 :

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

  2. $f/4$

  3. $2 f$

  4. $\text{infinite}$


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:

  1. 1 : 2 : 3

  2. 1 : 1 : 1

  3. 3 : 2 : 1

  4. 2 : 2 : 1


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:

  1. 20 cm

  2. 4 cm

  3. 12.5 cm

  4. 8 cm


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, $f = 12.5$

A convex lens has a focal length of $0.5 m$. It has to be combined with a second lens, so that the combination has a power of $1.5$ dioptre. Which of the following could be the second lens?

  1. A concave lens of focal length $2 m$

  2. Another convex lens of focal length $0.5 m$

  3. A concave lens of focal length $0.5 m$

  4. A convex lens of focal length $2 m$


Correct Option: A
Explanation:

Power of combination $P = 1.5 $ dioptre

Power of convex lens having focal length of $0.5 m$,  $P _1 = \dfrac{1}{0.5}$ dioptre
Using $P = P _1 + P _2$
$\therefore$ $1.5 = \dfrac{1}{0.5} + P _2$
We get $P _2 = -0.5$ dioptre
Thus focal length of second lens $f _2 = \dfrac{1}{P _2}$ 
$\implies$ $f _2 =- \dfrac{1}{0.5} = -2 m$ (negative sign implies the lens must be a concave lens)
Thus second lens could be concave lens of focal length $2m$.

A combination of convex and concave lenses has power $ 4 D $ .If the convex lens has power $5 D $ focal length of the concave lense will be 

  1. $ 100 cm$

  2. $ -100 cm$

  3. $-1 cm$

  4. $-\dfrac{100}{9}cm$


Correct Option: B
Explanation:

Power of combination 
$P=P _1+P _2$
$\Rightarrow   4=5+P _2$
$\Rightarrow  P _2=4-5=-1 D$
$\therefore   -1=P _2=\dfrac{100}{f(cm)}$
$\therefore   f=-100 cm$

Two identitical thin plano-convex glass lenses (refractive index 1.5 ) each having radius of curvature of 20 cm are placed with their convex surface in contact at the centre .The intervening space is filled with oil of refractive index 1.7.The focal length of the combination is 

  1. -20 cm

  2. -25 cm

  3. -50 cm

  4. 50 cm


Correct Option: C
Explanation:

The intervening space will act as concave lens.

Thus, the net focal length of the combination of three lenses would be given as, $\dfrac{1}{f} = \dfrac{1}{f _1}+\dfrac{1}{f _2}+\dfrac{1}{f _3}$

Now by the lens makers' formula,
$\dfrac{1}{f _1}=(1.5-1)\bigg(\dfrac{1}{\infty}-\dfrac{1}{-20}\bigg)=\dfrac{1}{40}$
$\dfrac{1}{f _2}=(1.7-1)\bigg(\dfrac{1}{-20}-\dfrac{1}{20}\bigg)=-\dfrac{2.8}{40}$
$\dfrac{1}{f _3}=(1.5-1)\bigg(\dfrac{1}{20}-\dfrac{1}{\infty}\bigg)=\dfrac{1}{40}$
$\therefore \dfrac{1}{f}=-\dfrac{0.8}{40}\implies f=-50cm$

Option C is correct.

A lens of power $6D$ is put in contact with a lens of power $-4\ D$. The combination will behave like a:

  1. Divergent lens of focal length $25\ cm$

  2. Convergent lens of focal length $50\ cm$

  3. Divergent lens of focal length $20\ cm$

  4. Convergent lens of focal length $100\ cm$


Correct Option: B
Explanation:

For a combination of lenses, net power of the combination is given by sum of individual powers i.e.

            Pnet = ${ P } _{ 1 }+{ P } _{ 2 }$
According to question
            Pnet = 6D - 4D = +2D
hence it will be convergent lens of focal length = $\dfrac { 100 }{ 2 } cm=50cm$