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CBSE Class 10 Answered

The power of the eye-lens of the normal eye where it is focused at its far point and at the near point respectively is  (Assume distance between retina from eye-lens to be 2.5cm) (PLEASE SOLVE IT IN DETAIL, especially in sign convention used here...)
Asked by anshoomaurya04 | 23 Feb, 2020, 11:09: PM
answered-by-expert Expert Answer
Object at far point
 
we have lens equation ,  (1/v) - (1/u) = 1/f  ...............(1)
 
v = lens-to-image distance = 2.5 cm , u = lens-to-object distance = -∞
 
Hence from eqn.(1), we get 1/f = 1/2.5 + (1/ ∞ ) , i.e., f = 2.5 cm
 
Power of lens = 1/f  , if focal length f is expressed in meter
 
Hence power of eye lens when focussing far point object = 1/ ( 0.025 ) = 40 diopter
---------------------------------------
 
For near point object,  u = -25 cm , v = 2.5 cm
 
(1/v) - (1/u) = 1/f
 
(1/2.5) + (1/25) = 1/f   or    f = 25/11 = 2.27 cm
 
power of eye lens = 1/0.0227≈ 44 diopter
 
 
Answered by Thiyagarajan K | 23 Feb, 2020, 11:45: PM

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