
For normal eyes, the cornea provides a converging power of \[40\text{ D}\] and the least converging power of the eye lens behind the cornea is \[20\text{ D}\]. Using this information, the distance between the retina and the cornea-eye lens can be estimated to be:
A. \[1.5\text{ cm}\]
B. \[5\text{ cm}\]
C. \[2.5\text{ cm}\]
D. \[\text{1}\text{.67 cm}\]
Answer
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Hint: The cornea and the eye lens behave as a combination of two thin converging lenses kept in contact, focusing light on the retina.
Formula used:
The power (in diopter) of two thin lenses in contact is given by
\[{{P}_{eff}}={{P}_{1}}+{{P}_{2}}\]
Where \[{{P}_{1}}\] is the power of the lens on which the light is first incident at, and \[{{P}_{2}}\] is the power of the second lens in the combination.
The effective focal length (in metre) of the equivalent lens is given by
\[{{f}_{eff}}=\dfrac{1}{{{P}_{eff}}}\]
Complete step by step solution:
The power or cornea, \[{{P}_{1}}=40\text{ D}\]
The power of eye-lens, \[{{P}_{2}}=20\text{ D}\]
The effective power of the combination of the two lens is
\[{{P}_{eff}}=40\text{ D}+20\text{ D}=60\text{ D}\]
The cornea-eye lens combination focuses the light on the retina.
So, the effective focal length is equal to the distance between the retina and the cornea-eye lens.
Now, the effective focal length is
\[\begin{align}
& {{f}_{eff}}=\dfrac{1}{{{P}_{eff}}} \\
& \text{ }=\dfrac{1}{60}=0.0167\text{ m}=1.67\text{ cm} \\
\end{align}\]
Therefore, option D. is the correct answer.
Additional information: Light enters the eye through the cornea, a curved front surface. It passes through the central hole in the iris called the pupil. The size of the pupil can change under control of muscles. The light is further focused by the eye lens on the retina. The retina contains rods and cones which are light and color sensitive cells respectively and transmits the electrical signals to the optic nerve.
The focal length of the eye-lens can be modified somewhat by the ciliary muscles. So, for objects at all distances, images are formed at the retina.
The closest distance for which the eye lens can focus light on the retina is called the least distance of distinct vision. The standard value for normal vision is taken as \[25\text{ cm}\].
Note: Power of a converging lens is positive because its focal length is positive.
To determine power of a lens, the focal length should be taken in metres.
Formula used:
The power (in diopter) of two thin lenses in contact is given by
\[{{P}_{eff}}={{P}_{1}}+{{P}_{2}}\]
Where \[{{P}_{1}}\] is the power of the lens on which the light is first incident at, and \[{{P}_{2}}\] is the power of the second lens in the combination.
The effective focal length (in metre) of the equivalent lens is given by
\[{{f}_{eff}}=\dfrac{1}{{{P}_{eff}}}\]
Complete step by step solution:
The power or cornea, \[{{P}_{1}}=40\text{ D}\]
The power of eye-lens, \[{{P}_{2}}=20\text{ D}\]
The effective power of the combination of the two lens is
\[{{P}_{eff}}=40\text{ D}+20\text{ D}=60\text{ D}\]
The cornea-eye lens combination focuses the light on the retina.
So, the effective focal length is equal to the distance between the retina and the cornea-eye lens.
Now, the effective focal length is
\[\begin{align}
& {{f}_{eff}}=\dfrac{1}{{{P}_{eff}}} \\
& \text{ }=\dfrac{1}{60}=0.0167\text{ m}=1.67\text{ cm} \\
\end{align}\]
Therefore, option D. is the correct answer.
Additional information: Light enters the eye through the cornea, a curved front surface. It passes through the central hole in the iris called the pupil. The size of the pupil can change under control of muscles. The light is further focused by the eye lens on the retina. The retina contains rods and cones which are light and color sensitive cells respectively and transmits the electrical signals to the optic nerve.
The focal length of the eye-lens can be modified somewhat by the ciliary muscles. So, for objects at all distances, images are formed at the retina.
The closest distance for which the eye lens can focus light on the retina is called the least distance of distinct vision. The standard value for normal vision is taken as \[25\text{ cm}\].
Note: Power of a converging lens is positive because its focal length is positive.
To determine power of a lens, the focal length should be taken in metres.
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