A little fish of 1 cm of height is at the center of a spheric fishbowl with diameter of 20 cm filled with water (n=1.33). The material of the sphere is so thin that its effects of refraction can be neglected. A thick equiconvex lens with thickness 2 cm and radii of curvature of 15cm and refraction index of 1.5 is located with its first vertex at a distance of 50cm of the center of the bowl. What is the position on the optic axis of the image of the fish? What is the size of the image ?

A biconcave lens has radii of curvature equal to R₁ = 31.0 cm and R₂ = 43.7 cm. A very distant object on the R₁ side of the lens forms an image. The lens medium has index of refraction 1.53 for violet light and 1.51 for red light. (a) Where is the image formed by violet light? (Enter your answer in cm and indicate the position of the image relative to the lens with the sign of your answer. Give your answer to at least two decimal places.) cm (b) Where is the image formed by red light? (Enter your answer in cm and indicate the position of the image relative to the lens with the sign of your answer. Give your answer to at least two decimal places.) cm

Compute the focal length of a diverging thin lens made of flint glass, whose refractive index is 1.66 and is immersed in air having refractive index 1. The radii of the spherical surfaces of the lens are 10 cm and 20 cm. Select one: O -30 cm O O -10 cm 30 cm 10 cm

A 96.0 cm long cylindrical stick is made of glass with refractive index 1.60. The ends of the stick have been polished to form convex spherical surfaces of radii 8.00 cm and 16.0 cm, respectively. An object is in the air about the symmetry axis of the stick 20.0 cm from the end that has a radius of 8.00 cm. (a) Determine the position of the final image due to refraction at both spherical surfaces. (b) Is the final image real or virtual? Assume that the rays are paraxial. Hint: The image of the first interface serves as the object for the second interface. (a) Inside the stick 84 cm from the object; (b) Virtual (a) Inside the stick 84 cm from the object; (b) Real (a) Inside the stick 64 cm from the object; (b) Virtual (a) Inside the stick 64 cm from the object; (b) Real (a) Inside the stick 44 cm from the object; (b) Virtual (a) Inside the stick 44 cm from the object; (b) Real (a) Inside the wand 36 cm from the object; (b) Virtual (a) Inside the stick 36 cm from the object; (b) Real…

not a graded assignment

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.…

A 1.00-cm-high object is placed 3.95 cm to the left of a converging lens of focal length 7.80 cm. A diverging lens of focal length –16.00 cm is 6.00 cm to the right of the converging lens. Find the position and height of the final image. position 7.5 cm in front of the second lens v 0.5466 height Calculate the magnification produced by each lens. Then consider how the magnification relates image size and object size for each lens to find the height of the final image. cm Is the image inverted or upright? upright O inverted Is the image real or virtual? real virtual

A light ray incident parallel to a biconvex lens made from glass (n 2 = 1.55) and has a radii of curvature (R1 = 12 cm; R2 = 8 cm). If the lens placed separates between the air (n 1 = 1.0) and liquid (n3 = 1.4), Where is the ray convergence in the liquid

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 technician in a pathology laboratory is using a compound microscope to examine tissue samples. The focal length of the objective of the microscope is 0.35 cm, and the focal length of the eyepiece is 2.5 cm. The two lenses are separated by 27 cm. One day, the technician forgot to wear his contact lenses, which have a prescription of +1.5 diopters and allow him to see objects clearly at 25 cm. When he looks into the microscope without wearing his contacts, what is the approximate angular magnification of the scope? Number Units