1.

Derive an expression for a thin double convex lens. Can you apply the same to a double concave lens too?

Answer»

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Solution :
i) convex lens is made up of two spherical refracting surfaces of radii of curvatures, `R_(1) and R_(2) and mu` is the refractive index of the lens.
ii) `P_(1),P_(2)` and the poles, `C_(1), C_(2)` are the centres of curvatures of two surfaces with optical centre C.
iii) Consider a point object O LYING on the principal axis of the lens and `I_(1)` is the real image of the object.
If `CI_(1)~~P_(1)I_(1)=v_(1)`
and `"CC"_(1)~~PC_(1)=R_(1)`
`CO~~P_(1)O=u`
iv) As refraction is taking place from rarer to denser medium
`(mu_(1))/(-u)+(mu_(2))/(v_(1))=(mu_(2)-mu_(1))/(R_(1))"..................................(1)"`
V) The refracted ray suffers further refraction
THEREFORE I is the final real image of O.
vi) For refraction at second surface, `I_(1)` as virtual object, whose real image is formed at I,
`therefore u~~CI_(1)~~P_(2)I_(1)=V_(1)`
`"Let "CI~~P_(2)I=V`
vii) Now refraction taking place from denser to rarer medium
`(-mu_(2))/(v_(1))+(mu_(1))/(v)=(mu_(1)-mu_(2))/(R_(2))=(mu_(2)-mu_(1))/(-R_(2))"...................(2)"`
Adding equations (1) and (2)
`(mu_(1))/(-u)+(mu_(1))/(v)=(mu_(2)-mu_(1))((1)/(R_(1))-(1)/(R_(2)))`
`mu_(1)=((1)/(v)-(1)/(u))=(mu_(2)-mu_(1))((1)/(R_(1))-(1)/(R_(2)))`
`(1)/(v)-(1)/(u)""=((mu_(2))/(mu_(1))-1)((1)/(R_(1))-(1)/(R_(2)))(because mu=(mu_(2))/(mu_(1)))`
`(1)/(v)-(1)/(u)""=(mu-1)((1)/(R_(1))-(1)/(R_(2)))"....................(3)"`
When the object on the left of the lens is at infinity `(PROP)`, image is formed at principal focus of the lens.
`therefore""u=prop, upsilon = f=" focal length of the lens."`
`(1)/(f)-(1)/(prop)=(mu-1)((1)/(R_(1))-(1)/(R_(2)))`
`(1)/(f)=(mu-1)((1)/(R_(1))-(1)/(R_(2)))"........................(4)"`
This is the lens marker's formula.
Yes, same formula applies to double concave lens too.


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