Question

The ciliary muscles of eye control the curvature of the lens in the eye and hence and alter the effective focal length of the system. When the muscles are fully relaxed,
the focal length is maximum. When the muscles are strained, the curvature of lens
increases. That means, the radius of curvature decreases and focal length decreases.
For a clear vision, the image must be on the retina. The image distance is therefore
fixed for clear vision and it equals the distance of the retina from the eye lens. It is about
$2.5cm$ for a grown up person.

A person can theoretically have clear vision of an object situated at any large
distance from the eye. The smallest distance at which a person can clearly see is
related to minimum possible focal length. The ciliary muscles are most strained in
this position. For an average grown up person, minimum distance of the object
should be around 25 cm.

A person suffering from eye defects uses spectacles (eye glass). The function of lens
of spectacles is to form the image of the objects within the range in which the
the person can see clearly. The image of the spectacle lens becomes object for the eye
lens and whose image is formed on the retina.

The number of spectacle lens used for the remedy of eye defect is decided by the
power of the lens required and the number of spectacle lens is equal to the
numerical value of the power of lens with sign. For example, if power of the lens
required is +3D (converging lens of focal length 100/3 cm), the number of lens
will be +3.

For all the calculations required, you can use the lens formula and lens maker's formula. Assume that the eye lens is an equiconvex lens. Neglect the distance between the eye lens and the spectacle lens.

Minimum focal length of eye lens of a normal person is
A.)25cm
B.)2.5cm
C.)25/9 cm
D.)25/11 cm

Answer 1

Hint: The focal length of our eye changes with object distance. The closer the object is to our eye, the more powerful the lens should be and hence, the lens has the smallest focal length when the object is as close as possible.

Complete step by step answer
We are given the question that the distance between the image and the lens is always a constant. When we focus on far objects, our lens is receiving parallel rays. But we know that parallel rays hitting a lens converge at the principal focus. If our lens focuses these parallel rays onto our retina, the focal length of our lens should be 2.5cm.

But when an object comes closer, two rays emerging from the object would be diverging. To focus such diverging rays at the same 2.5cm distance, we need a more powerful lens of lesser focal length.

So as we focus on objects that are closer and closer, the ciliary muscles contract reducing the focal length of the lens. When the object comes closer than 25 cm, our ciliary muscles cannot reduce the focal length anymore and we reach the minimum possible focal length.

This means that the smallest focal length of our eye lens is attained when the object is at 25 cm and the image is formed at 2.5cm.

Now we can use the lens formula to find the minimum focal length. If $u$ is the distance if an object from the lens and $v$ the distance of the image from the lens, then lens formula gives the focal length of lens $f$ as

$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

Applying sign conventions to object and image distance gives $u = - 25\;cm$ and $v = + 2.5cm$

We can substitute these values in (1) to get :

$\dfrac{1}{f} = \dfrac{1}{{2.5}} - \dfrac{1}{{ - 25}}$
$\dfrac{1}{f} = \dfrac{{10}}{{25}} + \dfrac{1}{{25}}$
$\dfrac{1}{f} = \dfrac{{11}}{{25}}$ or $f = \dfrac{{25}}{{11}}cm$
So the correct answer is option D.

Note: The sign conventions have to be followed while measuring $u$, $v$, and $f$ all distances measured to left are taken negative and those to the right are positive.
Also note that it is not only the lens in our eye that converges the rays. Two fluid filled chambers called aqueous humour and vitreous humour do most of the converging. The lens does only the fine adjustments so that the image is perfectly focused.