Question

Calculate the power of the eye- lens of the normal eye when it is focused at its
(a) Far point, infinity and
(b) Near point, \[25cm\] from the eye.
Assume the distance of the retina from the eye-lens to be \[2.5cm\].

Answer 1

Hint:-
- Type of lens present in the eye is a convex lens.
- Power of eye is the inverse of focal length.
- A convex lens image is formed at focus if the object is at far, infinity.
- Recall the lens equation \[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]

Complete step by step solution:-
Here, in the eye lens is a convex lens.
(a) Far point, infinity,
If the object is at infinity the image is obtained at retina.
So the distance between the eye lens and retina is the focal length.
It is already given,
So focal length \[f = 2.5cm = 0.025m\]
Power is ,
 \[\begin{gathered}
  P = \dfrac{1}{f} \\
  P = \dfrac{1}{{0.025}} = 40D \\
\end{gathered} \]

(b) Near point, \[25cm\] from the eye.
 This case is different from the initial problem, the object is at \[25cm\], here light is coming from object to lens.
But measurements are always from the lens. So according to sign conversion the distance of the object from the lens (\[u\]) is negative.
\[\begin{gathered}
  P = \dfrac{1}{f} = 0.44c{m^{ - 1}} = 44{m^{ - 1}} = 44D \\
  1D = 1{m^{ - 1}} \\
\end{gathered} \]
Image is obtained at retina, the measurement from the lens to retina is the same direction of the light ray. So \[v\] is positive. Distance between the retina and lens is already given,
\[v = 2.5cm\]
From the lens equation,
\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} = (\dfrac{1}{{2.5}}) - (\dfrac{1}{{ - 25}}) = \dfrac{1}{{2.5}} + \dfrac{1}{{25}}\]
\[\dfrac{1}{f} = \dfrac{{25 + 2.5}}{{25 \times 2.5}} = \dfrac{{27.5}}{{62.5}} = 0.44c{m^{ - 1}}\]
Power \[P = \dfrac{1}{f} = 0.44c{m^{ - 1}} = 44{m^{ - 1}} = 44D\]

So the answer is
(a) Far point, infinity,
Power \[P = 40D\]
(b) Near point, \[25cm\] from the eye.
Power \[P = 44D\]

Note:-
- We can use the conversion\[1cm = {10^{ - 2}}m = 0.01m\].
- We can recall the conversion\[1c{m^{ - 1}} = {10^2}{m^{ - 1}} = 100{m^{ - 1}}\].
- The unit for power of lens is Diopter (\[D\])
\[1D = 1{m^{ - 1}}\]
- A change in shape of the eye can alter the focal length of the eye.
- Sign convention always considers these types of problems.