Engineering
Physics
Lens
Question

The ciliary muscles control the curvature of the lens in the eye and hence can 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 the lens increases and the focal length decreases. For a clear vision, the image must be formed 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 2.5 cm for a grown – up person. If we apply the lens formula to eye, the magnitudes of the object – distance, the image – distance and the effective focal length satisfy.

1vo-1(-u0)=1f

Here, v0 is fixed, hence by changing f, the eye can be focused on objects placed at different distances u0. For an average grown – up person, u0 can be varied from around 25 cm or less to infinity for clear vision i.e. image focused on retina. Nearsightedness or Myopia. A person suffering from this defect cannot see distant objects clearly. This is because fmax is less than the distance from the lens to the retina and the parallel rays coming from the distant object focus short of the retina. The ciliary muscles are fully relaxed in this case and any strain in it can only further decrease the focal length which is of no help to see distant objects.

The remedy of myopia is quite simple. The rays should be made a bit divergent before entering the eye so that they may focus a little later. Thus, a divergent lens should be given to a myopic person to enable him/her to see distant objects clearly.

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Linked Question 1

Maximum power of eye lens is about

104D

44D

14D

4D

Solution
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Max. power ⟹ Power ⟹ least focal length

1v-1u=1f

12.5-1-25=1ff=2511 cm=144 m power =44D.

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Linked Question 2

The maximum focal length of the eye – lens of a person is greater than its distance from the retina. The eye is

strained for objects at short distances only

unstrained for all distances

always strained in looking at an object

strained for objects at large distances only

Solution
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Maximum focal length is attained if eye is fully relaned i.e. to focus seen the parellel rays eyes have to be stressed to decrease focal length.

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Linked Question 3

A normal eye is not able to see objects closer than 25 cm because

the eye is not able to decrease the focal length beyond a limit

the focal length of the eye is 25 cm

the eye is not able to decrease the distance between the eye – lens and the retina beyond a limit

the distance of the retina from the eye – lens is 25 cm

Solution
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Self enplainatory

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Linked Question 4

A nearsighted person cannot clearly see beyond 250 cm. Find the power of the lens needed to see objects at large distances.

– 0.4 D

2.5/11D

2.5/9 D

– 0.4 D

Solution
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If image of ∞ is formed at 250 cm lens works will

1-250-1=1f    f=-250 cm

f = – 2.5 m

P=1-2.5=-0.4D.

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Linked Question 5

When objects at different distances are seen by the eye, which of the following remain constant?

The image – distance from the eye – lens

The radii of curvature of the eye – lens

The focal length of the eye – lens

The object – distance from the eye – lens

Solution
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Self enplainatory

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