
A normal eye has retina \[2\,cm\] behind the eye-lens. What is the power of the eye-lens when the eye is fully relaxed and most strained?
Answer
479.1k+ views
Hint: We start by analyzing the given data. Then we move onto finding the focal length of the lens and the reciprocal of this value will be the power of the lens. then we move to find the value of focal length again but this time with respect to the mentioned change in the object distance.
Formulas used:
The power of a particular lens can be found out by the formula,
\[P = \dfrac{1}{f}\]
The lens formula gives us the relationship between the focal point, the image and the object distance, it is as follows,
\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]
Complete step by step answer:
We are approaching this question as two cases, case (a) for a fully relaxed eye and case (b) for a most strained eye. Going on to the cases,
Case(a): When the eye is fully relaxed,
When the eye lens is fully relaxed, we have \[u = \infty \]
Distance from the retina to the eye, \[v = 2 = 0.02\,m\]
Using the lens formula, we find the focal length as,
\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} \\
\Rightarrow \dfrac{1}{f} = \dfrac{1}{{0.02}} - \dfrac{1}{\infty } \\
\Rightarrow \dfrac{1}{f} = 50\]
We take the reciprocal to find the focal length but since the relation between focal length and power is, \[P = \dfrac{1}{f}\] , we have directly found out the power of the eye-lens when it is fully relaxed.
Therefore, the power of the eye-lens when it is fully relaxed will be \[50\,D\].
Case(b): When the eye-lens is mostly strained, the value of object distance will be \[u = - 25\,cm = - 0.25\,m\] because this is still where an average human being can see properly. The value of image distance will be the same because the retina does not move,
\[v = 2\,cm = 0.02\,m\]
Using the lens formula,
\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} \\
\Rightarrow \dfrac{1}{f} = \dfrac{1}{{0.02}} + \dfrac{1}{{0.25}} \\
\therefore \dfrac{1}{f} = 54\]
Again, we take the reciprocal to find the focal length but since the relation between focal length and power is, \[P = \dfrac{1}{f}\] , we have directly found out the power of the eye-lens when it is mostly strained.
Therefore, the power of the eye-lens when it is mostly strained will be \[54\,D\].
Note: The human eye is a part of the body responsible for vision. The retina is a part of the human eye that receives light from the lens and converts it into neural signals. The focal length of a human eye is controlled by the ciliary muscles. Ciliary muscles control the accommodation of viewing objects at varying distances by regulating the flow of aqueous humor.
Formulas used:
The power of a particular lens can be found out by the formula,
\[P = \dfrac{1}{f}\]
The lens formula gives us the relationship between the focal point, the image and the object distance, it is as follows,
\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]
Complete step by step answer:
We are approaching this question as two cases, case (a) for a fully relaxed eye and case (b) for a most strained eye. Going on to the cases,
Case(a): When the eye is fully relaxed,
When the eye lens is fully relaxed, we have \[u = \infty \]
Distance from the retina to the eye, \[v = 2 = 0.02\,m\]
Using the lens formula, we find the focal length as,
\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} \\
\Rightarrow \dfrac{1}{f} = \dfrac{1}{{0.02}} - \dfrac{1}{\infty } \\
\Rightarrow \dfrac{1}{f} = 50\]
We take the reciprocal to find the focal length but since the relation between focal length and power is, \[P = \dfrac{1}{f}\] , we have directly found out the power of the eye-lens when it is fully relaxed.
Therefore, the power of the eye-lens when it is fully relaxed will be \[50\,D\].
Case(b): When the eye-lens is mostly strained, the value of object distance will be \[u = - 25\,cm = - 0.25\,m\] because this is still where an average human being can see properly. The value of image distance will be the same because the retina does not move,
\[v = 2\,cm = 0.02\,m\]
Using the lens formula,
\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} \\
\Rightarrow \dfrac{1}{f} = \dfrac{1}{{0.02}} + \dfrac{1}{{0.25}} \\
\therefore \dfrac{1}{f} = 54\]
Again, we take the reciprocal to find the focal length but since the relation between focal length and power is, \[P = \dfrac{1}{f}\] , we have directly found out the power of the eye-lens when it is mostly strained.
Therefore, the power of the eye-lens when it is mostly strained will be \[54\,D\].
Note: The human eye is a part of the body responsible for vision. The retina is a part of the human eye that receives light from the lens and converts it into neural signals. The focal length of a human eye is controlled by the ciliary muscles. Ciliary muscles control the accommodation of viewing objects at varying distances by regulating the flow of aqueous humor.
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