
If the focal length of the human eye is 2 cm, then find the focal length of the contact lens such that a combined focus of $ 2.5{\text{ }}cm $ is obtained after using the contact lens.
Answer
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Hint: The power of a lens is the inverse of the focal length of the length. When two lenses are kept in contact, the net power of the combination is the sum of the individual power of the lenses.
Formula Used: In this solution we will be using the following formula,
$ {P_{combo}} = {P_1} + {P_2} $ where $ {P_1},{P_2} $ are the powers of the lenses.
$ \dfrac{1}{{{f_{combo}}}} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} $ where $ {f_1},{f_2} $ are the focal lengths.
Complete step by step answer
We’ve been given that an eye is in contact with another lens such that the focal length of the combination is $ 2.5{\text{ }}cm $. Using the relation of the power of a lens with its formula, we can determine the focal length of the combination.
$ {P_{combo}} = {P_1} + {P_2} $
Since the power of a lens is the inverse of the focal length, we can write
$ \dfrac{1}{{{f_{combo}}}} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} $
Substituting $ {f_1} = 2 $ and $ {f_{combo}} = 2.5 $, we get
$ \dfrac{1}{{2.5}} = \dfrac{1}{2} + \dfrac{1}{{{f_2}}} $
Taking the known terms on the left side, we get
$ \dfrac{1}{{{f_2}}} = \dfrac{1}{{2.5}} - \dfrac{1}{2} $
$ \dfrac{1}{{{f_2}}} = - \dfrac{{0.5}}{5} $
Multiplying the numerator and the denominator on the right side by 10, we get
$ \dfrac{1}{{{f_2}}} = - \dfrac{1}{{10}} $
Which gives us
$ {f_2} = - 10\,cm $
Hence the focal length of the second lens that is in contact with the eye will be -10 cm.
Note
Since the eye lens is convex in nature it will have a positive focal length while the lens in contact has a negative focal length. For two lenses in contact, it is their power that will add up and not their focal length so we must be careful of not adding the individual focal length to find the focal length of the combination. Here we have assumed that there is no space between the lens and the eye lens however in reality any kind of lens that we put to our eye will have some space between it and the eye lens.
Formula Used: In this solution we will be using the following formula,
$ {P_{combo}} = {P_1} + {P_2} $ where $ {P_1},{P_2} $ are the powers of the lenses.
$ \dfrac{1}{{{f_{combo}}}} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} $ where $ {f_1},{f_2} $ are the focal lengths.
Complete step by step answer
We’ve been given that an eye is in contact with another lens such that the focal length of the combination is $ 2.5{\text{ }}cm $. Using the relation of the power of a lens with its formula, we can determine the focal length of the combination.
$ {P_{combo}} = {P_1} + {P_2} $
Since the power of a lens is the inverse of the focal length, we can write
$ \dfrac{1}{{{f_{combo}}}} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} $
Substituting $ {f_1} = 2 $ and $ {f_{combo}} = 2.5 $, we get
$ \dfrac{1}{{2.5}} = \dfrac{1}{2} + \dfrac{1}{{{f_2}}} $
Taking the known terms on the left side, we get
$ \dfrac{1}{{{f_2}}} = \dfrac{1}{{2.5}} - \dfrac{1}{2} $
$ \dfrac{1}{{{f_2}}} = - \dfrac{{0.5}}{5} $
Multiplying the numerator and the denominator on the right side by 10, we get
$ \dfrac{1}{{{f_2}}} = - \dfrac{1}{{10}} $
Which gives us
$ {f_2} = - 10\,cm $
Hence the focal length of the second lens that is in contact with the eye will be -10 cm.
Note
Since the eye lens is convex in nature it will have a positive focal length while the lens in contact has a negative focal length. For two lenses in contact, it is their power that will add up and not their focal length so we must be careful of not adding the individual focal length to find the focal length of the combination. Here we have assumed that there is no space between the lens and the eye lens however in reality any kind of lens that we put to our eye will have some space between it and the eye lens.
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