SOLUTION SET UP The center of curvature of the first surface of the lens is on...
Consider a thin lens, but with different radius of curvature, denoted as R1, R2. The refractive index of the lens is n. (1) Derive analytically, when a monochromatic plane wave is shined normally on the lens. What would the transmitted complex wave function be? From the derived complex wave function, what is the focal length in terms of R1, R2, n? (2) Take R1 50cm, R2 35cm, n 1.5. If we place an object with height of 4cm in front...
A biconcave lens is made with glass (n 1.36) and has a magnitude of curvature T1 - 1.0 cm on the front surface and a magnitude of curvature T2-7.0 cm on the back surface, as pictured. r1 r2 This lens is in air and you may use the lensmaker's equation n- 1 to find other relationships in this problem, if needed. We place an object of height h 4.08 mm a distance s-6.0 cm from the center of the lens...
5. What was the focal distance (f) when the radius of curvature was 0.8 and index of refraction was 1.6? 6. Calculate the radius of curvature of a lens with a focal distance of 40.0 cm and an index of 1.2. 7. An object placed 3cm away from a lens projects a real image 0.6m behind the lens. What is this lens focal distance? 8. What is the lens magnification? 9. An object 20.0 cm to the left of a...
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...
o Consider a thin double concave lens, with each face having a radius of curvature of 35 cm. The glass composing the lens has an index of refraction of 1.4 1.What is the focal length of this lens? 2.If an 1 cm object is placed at 1.5x the focal length, where is the image and how big is it, according to the thin lens equation? 3.Is the image real or virtual? Right side up? Hint: when sketching your ray diagram,...
Exercise 9 A simple model of the human eye ignores its lens entirely. Most of what the eye does to light happens at the outer surface of the transparent cornea. Assume that this surface has a radius of curvature of 6.00 mm and that the eyeball contains just one fluid with a refractive index of 1.40 At what distance behind the cornea a very distant object will be imaged on the retina. Describe the image
For safety reasons, you install a rear-window lens with a -0.296 m focal length in your van. Before putting the van in reverse, you look through the lens and see the image of a person who appears to be 0.341 m tall and 0.235 m behind the van. Determine the following. (a) distance of the person behind the van -1.14 Can you write the thin lens equation which gives a relationship between the focal length of a lens, the object...
A typical human lens has an index of refraction of 1.430. The lens has a double convex shape, but its curvature can be varied by the ciliary muscles acting around its rim. At minimum power, the radius of the front of the lens is 10.0 mm, whereas that of the back is 6.00 mm. At maximum power, the radii are 6.50 mm and 5.50 mm, respectively. If the lens were in air, what would be the maximum power and associated...
A typical human lens has an index of refraction of 1.430. The lens has a double convex shape, but its curvature can be varied by the ciliary muscles acting around its rim. At minimum power, the radius of the front of the lens is 10.0 mm, whereas that of the back is 6.00 mm. At maximum power, the radii are 6.50mm and 5.5mm, respectively. If the lens were in air: What would be the maximum power and associated focal length...
A 4.0 cm tall object is 5.0 cm in front of a diverging lens with a focal length of -6.0 cm. A converging lens with a focal length of 6.0 cm is located 8.0 cm behind the diverging lens. (As viewed from the side, from left to right, the sequence is object - diverging lens - converging lens - observer. Rays then travel from left to right through the system.) (a) Use ray tracing to draw image 1 and image...