A student is reading a lecture written on a blackboard. The lenses in her eyes h

Question

A student is reading a lecture written on a blackboard. The lenses in her eyes have a refractive power of 60.00 diopters, and the lens-to-retina distance is 1.674 cm.

A student is reading a lecture written on a blackboard. The lenses in her eyes have a refractive power of 60.00 diopters, and the lens-to-retina distance is 1.674 cm. How far (in meters) is the blackboard from her eyes? If the writing on the blackboard is 4.80 cm high, what is the size of the image on her retina (including the proper algebraic sign)?

Explanation / Answer

A)The dioptic power is the inverse of the focal length f = 1/D = 1/60 = 0.01666 m or 1.66 cm

now 1/f = 1/p + 1/q

f= focal length, p = object diatnce , q = image distnce

1/1.66 = 1/p + 1/1.674

p = qf/(q-f)

p = 1.66*1.674/(1.674-1.66)

p = 380.16 cm = 3.80 m

Part B)

The magnification equation is M = -q/p = h'/h

h'/4.8 = -1.667/380.16

h' = -0.02104 cm

h' = 2.1*10^-2 m

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