Question

The cornea behaves as a thin lens of focal length approximately 1.80 {\rm cm}, although this varies a bit. The material of whichit is made has an index of refraction of 1.38, and its front surface is convex,with a radius of curvature of 5.00 {\rm mm}.(Note: The results obtained here are not strictlyaccurate, because, on one side, the cornea has a fluid with arefractive index different from that of air.)a) If this focal length is in air, what is the radius ofcurvature of the back side of the cornea? (in mm)b) The closest distance at which a typical person can focus onan object (called the near point) is about 25.0 {\rm cm}, although this varies considerably with age. Wherewould the cornea focus the image of an 10.0 {\rm mm}-tall object at the near point? (in mm)c) What is the height of the image in part B? (mm)d) Is this image real or virtual? Is it erect orinverted?

Answer 1

a) The radius of curvature of the back side of the cornea is approximately 0.969 mm. b) The cornea would focus the image of an object at the near point at a distance of approximately 55.6 mm. c) The height of the image is approximately -10.0 mm. d) The image formed by the cornea is real and inverted.

Step-by-step explanation:

a) To find the radius of curvature of the back side of the cornea, we can use the formula for thin lens:

1/f = (n-1)(1/R₁ - 1/R₂)

Since the front surface is convex with a radius of curvature of 5.00 mm and the index of refraction is 1.38, we can substitute these values and solve for R₂. Plugging in the values, we get:

1/1.80 = (1.38-1)(1/5.00 - 1/R₂)

Simplifying this equation gives us:

R₂ ≈ 0.969 mm

b) To find where the cornea would focus the image of an object at the near point, we can use the lens formula:

1/f = (n-1)(1/R₁ - 1/R₂)

Since the object distance is 25.0 cm and the focal length is known, we can rearrange the formula and solve for the image distance:

1/25.0 = (1.38-1)(1/5.00 - 1/R₂)

Simplifying this equation gives us:

R₂ ≈ 55.6 mm

c) To find the height of the image, we can use the magnification formula:

m = -di/do

Since the object height is known and the magnification is given as 1, we can rearrange the formula and solve for the image height:

hi = -mho

Substituting the values, we get:

hi = -1(10.0 mm)

hi = -10.0 mm

d) The image is real because it is formed by the cornea and lens system, and it is virtual because it cannot be projected onto a screen. The image height is negative, which means it is inverted.