Find focal lengths of lenses made of glass `(mu=3//2)` and placed in air.

Explanation

(i)

Given ,`R_(1)=10 cm,R_(2)=-20cm`
On applying Lens makers formula, we have
`1/f=(mu-1)(1//R_(1)-1//R_(2))=(3/2-1)(1/10-1/((-20)))=(1/2)(3/20)`
`therefore` Focal length of lens f= 40/3 cm
(ii)

Given,`R_(1)=oo,R_(2)=30cm`
From Lens markerl,s formula we have,
`1/f=(3/2-1)(1/(oo)-1/(-30))=1/60`
`therefore` Focal length f =60 cm
(iii)

Given,`R_(1)=-20 cm,R_(2)=60 cm`
From Lens Maker's formula, we have
`1/f=(3/2-1)(1/20-1/60)=(1/2)(2/60)`
`therefore` Focal length,
f=60 cm
(iv)

Given, `R_(1)=20 cm,R_(2)=20 cm`
From Lens Maker's formula, we have
`1/f=(3/2-1)(1/(-15)-1/15)=(1/2)((-2)/15)`