A. A converging lens forms an image of an 7.50 mm -tall real object. The image i

Question

A. A converging lens forms an image of an 7.50 mm -tall real object. The image is 11.5 cm to the left of the lens, 3.50 cm tall, and upright.Where is the object located?

B. The cornea of the eye has a radius of curvature of approximately 0.60 cm , and the aqueous humor behind it has an index of refraction of 1.35. The thickness of the cornea itself is small enough that we shall neglect it. The depth of a typical human eye is around 25.0 mm .What would have to be the radius of curvature of the cornea so that it alone would focus the image of a distant mountain on the retina, which is at the back of the eye opposite the cornea?

Explanation / Answer

A) height of the object = 7.5 mm

height of the image = 3.5 cm

location of the image = 11.5 cm to the left of lens

magnification = size of image / size of object = - (distance of image /distance of object)

= 35 / 7.5 = - (115) / Do

Do = -24.64 mm

the object is located 2.46 cm to the right of the lens

B) R = f(n2 - n1 ) /n2

R = (0.6/2) ( 1.35 -1 ) /1.35

R = 0.077 cm

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