
What is the power of combination of lenses?
Answer
408k+ views
Hint: Adding up the powers would give us the power of combination of lenses. You can calculate the $F$ (i.e.) focal length by \[\dfrac{1}{v} + \dfrac{1}{u}\] where $u$ is the object distance from lens and $v$ is image distance from lens.
Complete step-by-step solution:
Firstly, let us see what is power of a lens
The reciprocal of the focal length of a lens in metres is called the power of the lens
The capacity of a lens to bend the rays of light depends upon the focal length. The smaller the focal length, the greater is the bending of a ray of light and vice- versa. Thus the power of a lens to bend the rays of light is inversely proportional to the focal length of the lens.
Units of power of a lens: The units of the power of a lens are dioptres.
One dioptre is the power of a lens of one metre focal length. The power of a lens is denoted by D. It is this dioptric number which the doctors prescribe for the spectacles of a person.
Mathematically
D=1/Focal length in metre
=100/Focal length in centimetre
Here we would be using
\[\dfrac{1}{F} = \dfrac{1}{{{F_1}}} + \dfrac{1}{{{F_2}}}\] \[\]
Where, \[{F_1}\] and \[{F_2}\] are the focal lengths of the First, second lens respectively.
Then, \[P = {P_1} + {P_2}\] ,
Where, $P$ is the power of lens with focal length $F$, \[{P_1}\] is the power of lens whose focal length is \[{F_1}\],
\[{P_2}\] is the power of a lens whose focal length is \[{F_2}\] ..
Thus the power of combination of lenses is the algebraic sum of the powers of individual lenses.
If multiple lenses are separated by distance d metre between each of them, then power of combination of lenses is given by relation
\[P = {P_1} + {P_2} - d\left( {{P_1}{P_2}} \right)\]
Note: We can also calculate $F$ using lens maker’s formula
\[\dfrac{1}{F} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}} \right)\]
Where,
$F$= focal length
\[\mu = \] refractive index of the lens
\[{R_1} = \] Radius of the first lens
\[{R_2} = \] Radius of the second lens
Complete step-by-step solution:
Firstly, let us see what is power of a lens
The reciprocal of the focal length of a lens in metres is called the power of the lens
The capacity of a lens to bend the rays of light depends upon the focal length. The smaller the focal length, the greater is the bending of a ray of light and vice- versa. Thus the power of a lens to bend the rays of light is inversely proportional to the focal length of the lens.
Units of power of a lens: The units of the power of a lens are dioptres.
One dioptre is the power of a lens of one metre focal length. The power of a lens is denoted by D. It is this dioptric number which the doctors prescribe for the spectacles of a person.
Mathematically
D=1/Focal length in metre
=100/Focal length in centimetre
Here we would be using
\[\dfrac{1}{F} = \dfrac{1}{{{F_1}}} + \dfrac{1}{{{F_2}}}\] \[\]
Where, \[{F_1}\] and \[{F_2}\] are the focal lengths of the First, second lens respectively.
Then, \[P = {P_1} + {P_2}\] ,
Where, $P$ is the power of lens with focal length $F$, \[{P_1}\] is the power of lens whose focal length is \[{F_1}\],
\[{P_2}\] is the power of a lens whose focal length is \[{F_2}\] ..
Thus the power of combination of lenses is the algebraic sum of the powers of individual lenses.
If multiple lenses are separated by distance d metre between each of them, then power of combination of lenses is given by relation
\[P = {P_1} + {P_2} - d\left( {{P_1}{P_2}} \right)\]
Note: We can also calculate $F$ using lens maker’s formula
\[\dfrac{1}{F} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}} \right)\]
Where,
$F$= focal length
\[\mu = \] refractive index of the lens
\[{R_1} = \] Radius of the first lens
\[{R_2} = \] Radius of the second lens
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