A thin convex lens of refractive index 1.5cm has 20cm focal length in air. If the lens in completely immersed in a liquid of refractive index. 1.6, then its focal length will be

Options

A. `-160 cm`
B. `- 100 cm`
C. `+ 10 cm`
D. `+ 100 cm`

Answer

A

Explanation

From lens formula,
`(1)/(f)=(._(a)mu_(g)-1)((1)/(R_(1))-(1)/(R_(2)))`
`=(1.5-1)((1)/(R_(1))-(1)/(R_(2)))` ….(i)
Also, `._(l)mu_(g)=(mu_(g))/(mu_(l))=(1.5)/(1.6)`
`therefore " " (1)/(f')=(._(l)mu_(g)-1)((1)/(R_(1))-(1)/(R_(2)))`
`(1)/(f')=((1.5)/(1.6)-1)((1)/(R_(1))-(1)/(R_(2)))` ....(ii)
Dividing Eq. (i) by Eq. (ii), we get
`(f)/(f')=(((1.5)/(1.6)))/((1.5-1))=-(1)/(16xx0.5)`
`f'=-16xx0.5xx f=16xx0.5xx20`
`rArr " " f'=-160 cm`