Solution:For a normal sighted eye, near point is, d= 25 cmmagnification is, m= 6since, m=df where f is the focal lengthf=dm=256 cmNow when the near point is 50 cmthen, magnifying power, m=df=5025×6=12

Please find this...

S o l u t i o n : F o r a n o r m a l s i g h t e d e y e, n e a r p o i n t i s, d = 25 c m m a g n i f i c a t i o n i s, m = 6 \sin c e, m = \frac{d}{f} w h e r e f i s t h e f o c a l l e n g t h f = \frac{d}{m} = \frac{25}{6} c m N o w w h e n t h e n e a r p o i n t i s 50 c m t h e n, m a g n i f y i n g p o w e r, m = \frac{d}{f} = \frac{50}{25} \times 6 = 12

View Full Answer