Two rays traveling parallel to the principal axis strike a large plano-convex lens having a refractive index of 1.54. The convex face is spherical, with a radius of curvature of 21.6 cm, and the two rays are h1 = 0.300 cm and h2 = 6.90 cm from the principal axis. If we do not make the paraxial approximation, then these parallel rays will not actually cross the principal axis in exactly the same location (i.e., spherical aberration occurs). Calculate where ray 2 crosses the principal axis, measured relative to the lens's center of curvature. Calculate the difference in the positions where the two rays cross the principal axis. ( show all your work.) Assuming the paraxial approximation (equivalent to saying h is small compared to R), find an algebraic formula for the focal length of this lens. Your answer should depend only on n and/or R. f = (Hand in your derivation of this formula, showing all your work

an object of height 5 cm is placed 20 cm in front of a converging lens at focal length 10 cm. Behind the converging lens, and 25cm from it, there is a diverging lens of the focal length of 6 cm. Find the location of the final image, in centimeters, with respect to teh diverging lens. what is the magnification of the final image? what is the height of the final image?

A box has a mass of 2.50 kg is at rest on a horizontal frictionless surface. Four forces act on the box. The four forces are: F₁ = 20.0 N (North), F₂ = 30.0 N (East), F3 = 15.0 N (South), and F4 = 40.0 N (West). What is the resulting acceleration of the box? Select one: a. b. C. d. 4.47 m/² @ 26.6° (North of West) 31.3 m/2 @ 26.6° (West of North) 31.3 m/2 @ 26.6° (North of West) 4.47 m/2 @ 26.6° (West of North)

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…

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

A student stands dsub0=2.4 m in front of a floor-to-ceiling mirror. Her eyes are hsube=1.53 m above the floor and she holds a flashlight at a distance hsubf=.55 m above the floor. Calculate the angle in degrees that the flashlight makes with respect to the floor if the light is reflected into her eyes.

A horizontal light ray is incident on a crown glass prism as shown in the figure where β = 27.8°. Find the angle of deviation δ of the ray—the angle that the ray emerging from the prism makes with the incident ray.