
For a normal eye, the cornea provides a converging power of 40 D and the least converging power of the eye lens behind the cornea is 20D. Using this information, the distance between the retina and the cornea eye lens can be estimated to be,
A. 1.5 cm
B. 5 cm
C. 2.5 cm
D. 1.67 cm
Answer
594.3k+ views
Hint: Total power of the two lenses would give the net power of the eye. The human retina is a screen on which the images are formed so the distance between the eye lens and the human eye would be the focal length due to the effective power of the two lenses.
Complete step by step answer:
Power of a lens is the ability of a lens to converge a beam of light falling on the lens. It is the reciprocal of focal length of the lens.
The Si unit of power is Dioptre, 1 Dioptre is the power of a lens of focal length one metre.
\[\text{P(dioptre)=}\dfrac{\text{100}}{\text{f(cm)}}\]
For a combination of more than one lens the total power is the algebraic sum of individual powers of the lenses i.e.
\[P={{P}_{1}}+{{P}_{2}}\]
Therefore, the effective power of cornea lens and the lens behind the cornea is:
\[P=40D+20D=60D\]
Now, the distance between retina and cornea lens would be the focal length due to effective power of both the lenses.
From the above formula, focal length is given by:$\begin{align}
& f=1.67cm \\
& \\
\end{align}$
$f=\dfrac{100}{P}$
$f=\dfrac{100}{60}$
$f=1.67cm$
So, the distance between the retina and the cornea eye lens is 1.67cm
Thus, the correct answer is option D. 1.67cm
Note: Students must remember that the formula for effective power is used only when the lenses are in contact with each other. And, if the net power is positive the system of lenses behaves as a convex lens and if negative they behave as a concave lens.
Complete step by step answer:
Power of a lens is the ability of a lens to converge a beam of light falling on the lens. It is the reciprocal of focal length of the lens.
The Si unit of power is Dioptre, 1 Dioptre is the power of a lens of focal length one metre.
\[\text{P(dioptre)=}\dfrac{\text{100}}{\text{f(cm)}}\]
For a combination of more than one lens the total power is the algebraic sum of individual powers of the lenses i.e.
\[P={{P}_{1}}+{{P}_{2}}\]
Therefore, the effective power of cornea lens and the lens behind the cornea is:
\[P=40D+20D=60D\]
Now, the distance between retina and cornea lens would be the focal length due to effective power of both the lenses.
From the above formula, focal length is given by:$\begin{align}
& f=1.67cm \\
& \\
\end{align}$
$f=\dfrac{100}{P}$
$f=\dfrac{100}{60}$
$f=1.67cm$
So, the distance between the retina and the cornea eye lens is 1.67cm
Thus, the correct answer is option D. 1.67cm
Note: Students must remember that the formula for effective power is used only when the lenses are in contact with each other. And, if the net power is positive the system of lenses behaves as a convex lens and if negative they behave as a concave lens.
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