Figure(a) shows the basic structure of a human eye. Light refracts into the eye through the cornea and is then further redirected by a lens whose shape (and thus ability to focus the light) is controlled by muscles. We can treat the cornea and eye lens as a single effective thin lens.A “normal” eye can focus parallel light rays from a distant object O to a point on the retina at the back of the eye, where the processing of the visual information begins. As an object is brought close to the eye, however, the muscles must change the shape of the lens so that rays form an inverted real image on the retina. (a) Suppose that for the parallel rays of Figures, the focal length f of the effective thin lens of the eye is 2.50 cm. For an object at distance p = 40.0 cm, what focal length f' of the effective lens is required for the object to be seen clearly? (b) Must the eye muscles increase or decrease the radii of curvature of the eye lens to give focal length f' ?

Foundation

Physics Foundation

Human Eye And Colorful World

refraction from lens

Question

Figure(a) shows the basic structure of a human eye. Light refracts into the eye through the cornea and is then further redirected by a lens whose shape (and thus ability to focus the light) is controlled by muscles. We can treat the cornea and eye lens as a single effective thin lens.A “normal” eye can focus parallel light rays from a distant object O to a point on the retina at the back of the eye, where the processing of the visual information begins. As an object is brought close to the eye, however, the muscles must change the shape of the lens so that rays form an inverted real image on the retina. (a) Suppose that for the parallel rays of Figures, the focal length f of the effective thin lens of the eye is 2.50 cm. For an object at distance p = 40.0 cm, what focal length f' of the effective lens is required for the object to be seen clearly? (b) Must the eye muscles increase or decrease the radii of curvature of the eye lens to give focal length f' ?

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(a) When the eye is relaxed, its lens focuses on faraway objects on the retina, a distance i behind the lens. We set p = ∞ in the thin lens equation to obtain 1/i = 1/f, where f is the focal length of the relaxed effective lens. Thus, i = f = 2.50 cm. When the eye focuses on closer objects, the image distance i remains the same but the object distance and focal length change. If p is the new object distance and f ' is the new focal length, then , object distance u = –p using lens equation

\frac{1}{u} + \frac{1}{v} = \frac{1}{f}

\frac{1}{p} + \frac{1}{i} = \frac{1}{f}

We substitute i = f and solve for f ':

f = \frac{pf}{p + f} = \frac{(40.0 cm) (2.50 cm)}{(40.0 cm) + (2.50 cm)} = 2.35 cm

(b) Consider the lens maker’s equation

\frac{1}{f} = (n - 1) (\frac{1}{r_{1}} - \frac{1}{r_{2}})

where r₁ and r₂ are the radii of curvature of the two surfaces of the lens and n is the index of refraction of the lens material. For the lens pictured in Figure, r₁ and r₂ have about the same magnitude, r₁ is positive, and r₂ is negative. Since the focal length decreases, the combination (1/r₁) – (1/r₂) must increase. This can be accomplished by decreasing the magnitudes of both radii.