Unformatted text preview: BE 518 HOMEWORK 5 Due Tuesday Oct. 10 Problem 1. Above is a photo of an airplane through a window containing water droplets (which are roughly hemispherical). a. Using equations, explain why the images of the airplane appear inverted relative to the airplane in the background b. Using equations, explain why the size of each image appears to increase with the size of the droplet it’s in. a.Problem 2. A myopic person has wellcorrected vision with an eyeglass prescription of 2 Water refractive index n ≈ 4/3. A hemispherical droplet with radius R diopters. While removing his glasses he moves them 5 cm away from his eyes (along the has a focal length 1/f = (n1)/R ≈ 1/(3R) via the thick lens equation, so f direction of vision). What factor change, if any, will he see in the angular size of a distant ≈ 3R. Since object distance is very large, si = f = 3R. Transverse object? magnification M = s /s = 3R/s < 0. Inverted real images. b. T i o o MT proportional to R, so image size is proportional to droplet size 1 airplane in the background b. Using equations, explain why the size of each image appears to increase with the size of the droplet it’s in. Problem 2. A myopic person has wellcorrected vision with an eyeglass prescription of 2 diopters. While removing his glasses he moves them 5 cm away from his eyes (along the direction of vision). What factor change, if any, will he see in the angular size of a distant object? 6 The lenses have a focal length f = 50 cm. They create a virtual image of a
distant object at a distance 50 cm to the left of the glasses. When the lens
moves 5 cm to the left, so does the virtual image, now at 55 cm to the left of the
observer. Since the size of the virtual image does not change, the ratio of the
angular size is equal to the inverse ratio of the distances to the virtual images:
Angular size (shifted) / angular size (original) = distance (original) / distance
(shifted) = 50 cm / 55 cm = 10/11 = 0.909 Angular size becomes smaller, to 90.9% of the original
! 2 Problem 3. Find the effective focal lengths of the two lens combinations shown above. Each consists of two hemispherical glass (n=1.5, radius R) lenses placed in contact with one another. Which is more powerful? Problem 4 Show that the principal planes of any symmetric biconvex glass lens (with n=1.5) are A: The only surfaces with optical power are the two curved surfaces. Each located approximately 1/3 and 2/3 of the way through the lens. Assume the thickness of the has power D = (n1)/R = 0.5 / R
lens is much smaller than the radius of curvature of the curved surfaces. Placed together (with no gap) the powers of the two surfaces simply add: D total = 2 * 0.5/R = 1/R f = R Problem 5. Depth of focus. You’re imaging a fluorescent particle (point source) on an infinityB is just a sphere. From the thick lens equation,
corrected microscope (tube lens focal length fT) using an objective with magnification M and Effective focal length 1/f = (n1)[ 1/R + 1/R – (n1)*2 / (nR) ]
numerical aperture NA. The particle is displaced from the object focal plane by a small distance = (0.5)[[ 2/R –
2/(3R)] = 2/(3R)
d. How large a spot does the out of focus point create at the image plane (camera)? f = 3R/2 The combination in A is more powerful (has shorter focal length) 3 Show that the principal planes of any symmetric biconvex glass lens (with n=1.5) are located approximately 1/3 and 2/3 of the way through the lens. Assume the thickness of the lens is much smaller than the radius of curvature of the curved surfaces. 4 Depth of focus. You’re imaging a fluorescent particle (point source) on an infinity corrected microscope (tube lens focal length fT) using an objective with magnification M and numerical aperture NA. The particle is displaced from the object focal plane by a small distance d. How large a spot does the out of focus point create at the image plane (camera)? 2. 5 ...
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 Fall '17
 focal length, angular size

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