Question

A certain child's near point is 15.0 cm; her far point (with eyes relaxed) is 129 cm. Each eye lens is 2.00 cm from the retina. (a) Between what limits, measured in diopters, does the power of this lens-cornea combination vary? lower bound diopters upper bound diopters (6) Calculate the power of the eyeglass lens the child should use for relaxed distance vision. diopters Is the lens converging or diverging? O converging diverging

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Answer 1

For viewing an object at near point,

object distance = u = - 15 cm,

image distance = v = + 2 cm.

Hence, the focal length f of the eye lens is given by :

1 / f = 1 / v - 1 / u = ( u - v ) / uv = { ( - 15 - 2 ) / ( - 15 x 2 ) } cm^-1

or, 1 / f = 0.567 cm^-1 = 56.7 m^-1.

Hence, power of the lens for near vision is 56.7 diopters.

For viewing an object at far point,

object distance = u = - 129 cm,

image distance = v = + 2 cm.

Hence, the focal length f of the eye lens is given by :

1 / f = 1 / v - 1 / u = ( u - v ) / uv = { ( - 129 - 2 ) / ( - 129 x 2 ) } cm^-1

or, 1 / f = 0.508 cm^-1 = 50.8 m^-1.

Hence, power of the lens for far vision is 50.8 diopters.

So, upper bound = 56.7 diopters,

and, lower bound = 50.8 diopters.

For relaxed distance vision,

object distance = u = infinity,

image distance = v = - 129 cm ( since, the far point of the eye lens is 129 cm ),

Hence, the focal length f of the eyeglass lens is given by :

1 / f = 1 / v - 1 / u = ( 1 / - 129 cm ) - 0

or, 1 / f = 0.008 cm^-1 = 0.8 m^-1.

So, power of the eyeglass lens needs to be 0.8 diopters.

Since, power of the eyeglass lens is positive, the lens must be a diverging lens.