The lens at the front of your eye (index of refraction n l = 1.40) is surrounded

Question

The lens at the front of your eye (index of refraction nl = 1.40) is surrounded by muscles that can change its focal length. These adjustments are done so that the image is projected on the photoreceptors in the retina at the back of the eye. (see figure 1)

1) Is the image real or virtual? Explain.

2) If the retina is 2.0 cm behind the lens, what is the range of focal lengths required to see objects that are both 10 cm (4 inches) from the eye and very far away.

3)How does the magnification of the object depend on its distance from the eye? Explain.

Explanation / Answer

u is the distance of object from the lens

v is the distance of image from lens.

1.) We have a lens, the image is falling back , so it is real image.

2.) we have v = 2 cm

u = 10 cm

we have for lens,

1/f = 1/u + 1/v

1/2 + 1/10 = 1/f

so , f = 1.6666666666666667 cm

so focal length has to be less than 1.6666666666666667 cm

so range of focal lengths is 0 < f < 1.6666666666666667 cm

3)

we have v = 2cm which is fixed

magnification = m = - v / u = -2 /u ,

the higher the value of u lower is the value of magnification.

the distant the object (u is high ) is from the lens , the magnification would be low.

The more distant the object is from eye the smaller would it appear.

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