Now w apply the tinens equation the eye. When ight enters your eye, most of te f

Question

Now w apply the tinens equation the eye. When ight enters your eye, most of te focusing happens at heinmetace between hear and he oomea (he outermost element of eye) The eye also has a double oomvea lying behind he comea,that tompletes the of forming an image on resna The lens of four It gets rounder for near vision end Meter tor wson)The crystalina lens has an index of refraction or about (an For the lens shown (FigMm1). find the length. you oould consider this lensin isolation thom the what he image befor an olject min front of SOLUTION ourvature for the second surface is not on the of the surface of the lens is on the outgoing sde so Ru 460mm. SETUP The center of We solve for f and then use the resut inthethn-ens equation outgoing de Rh 5.6 mm 7,2 mm Part The object dslanoe 020 m 200 REFLECT The image sighly father from the than would befor an inhntely distant object. As expected for a converging lens, the focal length is posilwe Practice Problem: Suppose that no observe an object that is 500 m away telensin your eye fanens out so that tacts ke the crystaline lens in the example, exorp with radi of and Ch 810 How fa behind the lens would the image form in this case? Express your answer to three signiteantfigures and pay attention tounts. 5.96 Incorrect Try Again: 2 attempts remaining Continue

Explanation / Answer

i thik in the problem radius of curvature are given

1/f = (n-1) [1/R1 - 1/R2]

= (1.40 - 1) [ 1/10mm - 1/-9.10]

or, f = 11.91 mm

now, 1/di + 1/do = 1/f

or, 1/di + 1/5930 = 1/11.91

or, 1/di = 1/11.91 - 1/5930

or, di = 11.93mm ...........................................ans

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