4
$\begingroup$

To see near objects,the ciliary muscles contract and the suspensary ligaments loosen,making the lense thicker. İ don't understand how that helps us to see distant objects .

To see distant objects the ciliary muscles relax stretching the suspensary muscles making the lense thinner.

İ haven't been able to find information about how this helps us to see , why it happens.

$\endgroup$
0

2 Answers 2

7
$\begingroup$

A thinner lens has a greater focal length, hence rays get less refracted. Inversely it is for a thicker lens. For a far object the rays are almost parallel. Therefore they do not have to get refracted so much. But for a near object the rays are more divergent. Hence you need a smaller focal length in order to refract the rays as much that you get a sharp picture. See the picture below for the ray paths. Take also a look at Lensmaker‘s equation describing how the focal length depends on the radius of the lens. Ray path in the eye

Source: https://courses.lumenlearning.com/physics/chapter/26-1-physics-of-the-eye/

$\endgroup$
2
  • $\begingroup$ Thank you. How does a thinner lens have a greater focal length i would hace thought it's a shorter focal length. $\endgroup$
    – user57928
    Commented Sep 10, 2018 at 16:19
  • 3
    $\begingroup$ @user57928 You‘re welcome. A thinner lens refracts rays less than a thick lens. This means parallel rays will meet further away. Thus, the focal length is greater. Quantitively this is described by Lensmaker‘s equation. $\endgroup$ Commented Sep 10, 2018 at 16:23
1
$\begingroup$

The eye has two lenses. Most of the power is due to the cornea lens that is constant focal length. The inner lens is deformable and can change its power.

The muscles deform the lens curvature. This changes the power D=1/f of the lens and thus the visual system can focus far and near objects. $$D_i=\frac{n-n_0}{R_i}$$ The more curved surface will have smaller radius R and higher power or smaller focal length.

Also the power changes a bit due to the thickness d as $$D=D_1+D_2-\frac{d}{n}D_1D_2=\frac{1}{f}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.