The near point of an eye is 70.0 cm. (a) What should be the power of a correctiv

Question

The near point of an eye is 70.0 cm.

(a) What should be the power of a correctivelens prescribed to enable the eye to see an object clearly at24.0 cm?
P = 1 diopters

(b) If, using the corrective lens, the person can see an objectclearly at 25.0 cm but not 24.0 cm, by how many diopters did thelens grinder miss the prescription?

P = 2diopters too low (a) What should be the power of a correctivelens prescribed to enable the eye to see an object clearly at24.0 cm?
P = 1 diopters

(b) If, using the corrective lens, the person can see an objectclearly at 25.0 cm but not 24.0 cm, by how many diopters did thelens grinder miss the prescription?

P = 2diopters too low

Explanation / Answer

a) given object distance p = 24.0cm image distance is q = -70.0cm here -ve sign is used becausethe image must be virtual & upright. we have 1/p +1/q = 1/f or f = pq/q+p      = -24*70 / -70+24      = 36.52cm so the power of the lens is     P1 = 1/f          =  1/0.3652m           =2.738 diopters b) here p = 25cm & q = -70.0cm we have 1/p +1/q = 1/f or f = pq/q+p      = -25*70 / -70+25      = 38.88cm so the power of the lens is     P2 = 1/f          =  1/0.3888m           =2.572 diopters the error in the power of the lens is P = P1-P2         = 2.738diopters -  2.572 diopters         = 0.166 dipoterstoo low.         = 2.738diopters -  2.572 diopters         = 0.166 dipoterstoo low.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---