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1 p. 1 q. 1 f. = +. Is the eye depicted above nearsighted or farsighted ? 1) nearsighted 2) farsighted 3) could be either. An eyeglass prescription calls for lens powers of +2.1 D for the left eye and +2.5 D for the right eye .
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1 p 1 q 1 f = + Is the eye depicted above nearsighted or farsighted? 1) nearsighted 2) farsighted 3) could be either
An eyeglass prescription calls for lens powers of +2.1 D for the left eye and +2.5 D for the right eye. The positive power means the lens is converging. This person must be A) nearsighted B) farsighted C) cannot tell from information given
A) nearsighted Were the object moved even further back, its rays to the lens would become more parallel, focusing an image even closer to the lens’ focal point (moving the image closer to the focus…further from the retina. As the object is moved closer to the eye, the convergence point of the rays moves back (toward the retina). So this eye can focus much better on close objects, but be unable to focus distant objects. B) farsighted A positive lens power means the lens is convex, or converging. So without the lens, the rays must converge behind the retina, and so the person is farsighted.
p = 12 cm f1= 8 cm f2= 12 cm 40 cm
p = 12 cm f1= 8 cm p2 = 40 cm – q1 = 16 cm f2= 12 cm 40 cm Note: overall M = M1M2 = 6
p = 12 cm f1= 8 cm p2 = 32 cm – 24 cm f2= 12 cm Move the lenses closer together so that the 1st image is within the 2nd’s focal point. 32 cm
p = 12 cm f1= 8 cm p2 = 32 cm – 24 cm f2= 12 cm Move the lenses closer together so that the 1st image is within the 2nd’s focal point. 32 cm Note: overall M = M1M2 = -6
p = 12 cm f1= 8 cm p2 = 16 cm – 24 cm f2= 12 cm What if you move the lenses so close together the 2nd interrupts the formation of the 1st image? 16 cm The object is virtual!
p = 12 cm f1= 8 cm p2 = 16 cm – 24 cm f2= 12 cm What if you move the lenses so close together the 2nd interrupts the formation of the 1st image? 16 cm The object is virtual! Note: overall M = M1M2 = 1.2
A telescope has fo=985 mm and fe = 50 mm. Find the angular magnification of the telescope and its appropriate length. (a) (b)
The objective of a compound microscope has a focal length of fo = 0.40 cm while that of the eyepiece is fe = 3.0 cm. They are separated by L = 20.0 cm. A person with a normal eye (N = 25 cm) uses the microscope. • Find the angular magnification. • b)What is the greatest angular magnification • achievable using the eyepiece alone (as a • magnifying glass)?