In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the refraction occurs at the cornea, whose vertex is 2.60 cm from the retina. What should be the radius of curvature of the cornea such that the image of an object 40.0 cm from the cornea's vertex is focused on the retina?

It is your first day at work as a summer intern at an optics company. Your supervisor hands you a diverging lens and asks you to measure its focal length. You know that with a converging lens, you can measure the focal length by placing an object a distance ss to the left of the lens, far enough from the lens for the image to be real, and viewing the image on a screen that is to the right of the lens. By adjusting the position of the screen until the image is in sharp focus, you can determine the image distance s′s′ and then use the equation 1s+1s′=1f1s+1s′=1f, to calculate the focal length ff of the lens. But this procedure won'’t work with a diverging lens−−by itself, a diverging lens produces only virtual images, which can'’t be projected onto a screen. Therefore, to determine the focal length of a diverging lens, you do the following: First you take a converging lens and measure that, for an object 20.0 cmcm to the left of the lens, the image is 29.7 cmcm to the right of the lens.…

The crystalline lens of the human eye is a double-convex lens made of material having an index of refraction of 1.44 (although this varies). Its focal length in air is about 8.00 mm, which also varies. We shall assume that the radil of curvature of its two surfaces have the same magnitude. (Note: The results obtained in the parts A, B and C are not strictly accurate, because the lens is embedded in fluids having refractive indexes different from that of air.) For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of The eye's lens. Part A Find the radii of curvature of this lens. Express your answer in millimeters. R= 7.0 mm Submit ✓ Correct Set Up: = (n − 1)(-). If R is the radius of the lens, then R₁ = R and R₂=-R₁ ² + ² = }· m = ² = -² Solve: = (n − 1) (₁ - ₁) = (n − 1) ( ²2 - ) = ²(n-¹) R=2(n-1)f=2(1.44) (8.00mm) = 7.04mm Part B If an object 12.0 cm tall is placed 26.0 cm from the eye lens, where would the lens focus it? Express your answer in…

A thin lens is comprised of two spherical surfaces with radii of curvatures of 32.4 cm for the front side and -26.9 cm for the back side. The material of which the lens is compose has an index of refraction of 1.49. What is the magnification of the image formed by an object placed 43.7cm from the lens?

The crystalline lens of the human eye is a double- convex lens made of material having an index of refraction of 1.44 (although this varies). Its focal length in air is about 8.00 mm, which also varies. We shall assume that the radii of curvature of its two surfaces have the same magnitude. (Note: The results obtained in the parts A, B and C are not strictly accurate, because the lens is embedded in fluids having refractive indexes different from that of air.) Find the radii of curvature of this lens. R = ΜΕ ΑΣΦ Submit Request Answer Part B ? mm If an object 10.0 cm tall is placed 33.0 cm from the eye lens, where would the lens focus it? ΜΕ ΑΣΦ ? s' = Submit Request Answer Part C How tall would the image be? Π ΑΣΦ 3= Submit Request Answer ? mm mm