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Question: Two lenses of focal length \({{\text{f}}_1} = 10cm\) and \({{\text{f}}_2} = 20cm\) are kept as shown...

Two lenses of focal length f1=10cm{{\text{f}}_1} = 10cm and f2=20cm{{\text{f}}_2} = 20cm are kept as shown. The resultant power combination will be:

\eqalign{ & {\text{A}}{\text{. - }}10D \cr & {\text{B}}{\text{. }}5D \cr & {\text{C}}{\text{. - }}5D \cr & {\text{D}}{\text{. }}10D \cr}

Explanation

Solution

Hint: As we know, the power of a lens is defined as the reciprocal of the wavelength, i.e.P = 1f{\text{P = }}\dfrac{1}{f} . The unit of power of a lens is diopters (D). Here we find the power of the combination of a lens is the algebraic sum of the powers of the individual lenses in contact.
1F=1f1+1f2+........\therefore \dfrac{1}{F} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} + ........
Where f is a focal length.
P = P1+P2+....\Rightarrow {\text{P = }}{{\text{P}}_1} + {{\text{P}}_2} + ....

Complete step by step answer:
According to a given diagram
The focal length of the convex lens = 10cm10cm
The focal length of the concave lens =20cm - 20cm
And the focal length of the plane mirror =\infty
So, the resultant power of the combination of lens
1F=1f1+1f2+1f3\dfrac{1}{{\text{F}}} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} + \dfrac{1}{{{f_3}}} 1F=1101201 \Rightarrow \dfrac{1}{{\text{F}}} = \dfrac{1}{{10}} - \dfrac{1}{{20}} - \dfrac{1}{\infty }

1F=120\therefore \dfrac{1}{{\text{F}}} = \dfrac{1}{{20}}
F = 20cm = 0.2m\Rightarrow {\text{F = 20cm = 0}}{\text{.2m}}
Now the power of combination P = 1F{\text{P = }}\dfrac{1}{{\text{F}}}
P = 10.2=5D\Rightarrow {\text{P = }}\dfrac{1}{{0.2}} = 5{\text{D}}

Additional information: 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 beam 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.

Note: If two lenses are separated by distance d meter, then the power of a combination of lenses is given by the relation
P = P1+P2d.(P1.P2){\text{P = }}{{\text{P}}_1} + {{\text{P}}_2} - {\text{d}}{\text{.(}}{{\text{P}}_1}{\text{.}}{{\text{P}}_2}) Here d is the distance.
Always remember the power of a concave lens is negative, and the power of a convex lens is positive. We use a combination of lenses for higher magnification because lens combination multiplies each other's magnification power.