A 5.0 cm long needle is placed vertically at a distance 20 cm in front of a double convex lens made of a material of refractive index 1.5 having radii of curvature as 20 cm and 30 cm .Find the height of image formed.

Explanation

Size of nedle `=O=+2.0cm`
`u=-48cm`
`R_(1)=+20cm, R_(2)=-30cm`
`""^(a)mu_(g)=1.5`
`1/f=(""^(a)mu_(g)-1)(1/(F_(1))-1/(R_(2)))`
`1/f=(1.5-1)(1/20-1/(-30))`
`=0.5(1/20+1/30)`
`1/f=0.5((3+2)/60)=0.5(1/12)=1/24`
`impliesf=24cm`
By lens formula we have
`1/v-1/u=1/f`
`1/v-1/(-48)=1/24`
`1/v+1/48=1/24`
`implies1/v=1/24-1/48=(2-1)/48=1/48`
`impliesv=48cm`
Magnification `=m=I/O=v/u`
`I=v/u(O)=48/(-48)(2)`
`=-2cm`
`:.` Size of image of needle `I=-2cm`