A thin glass (refractive index 1.5) lens has optical power of `-5D` in air. Its optical power in a liquid medium with refractive index 1.6 will be

Options

A. 1D
B. `-1D`
C. 25D
D. 20

Answer

A

Explanation

As we know that, from Lens maker's formula,
`(1)/(f)=((mu_(2))/mu_(1)-1)((1)/(R_(1))-(1)/(R_(2)))`
As, `(1)/(f)`=power of lens(P)
where, f=focal length
`mu_(2) and mu_(1)`= refreactive index of 2nd and 1st medium
`R_(2) and R_(2)`= radius of curvature of 1st and 2nd curved surface of lens
`"So, for 1st case", P_(1)=((1.5)/(1)-1)((1)/(R_(1))-(1)/(R_(2))) ....(i)`
As, `mu_(2)`=1.5, mu=1(air)
Where `P_(1)`= power in initial condition
Now, for 2nd case, `P_(2)=((1.5)/(1.6)-1)((1)/(R_(1))-(1)/(R_(2))) ....(iI)`
Where, `P_(2)`=power in 2nd condition
On dividing Eqs. (i) and (ii), we get
`P_(1)/P_(2)=((1.5-1)/1)/((1.5-1.6)/1.6)=(0.5xx1.6)/(-0.1)rArr(-5)/P_(2)=(-0.8)/0.1(asP_(1)=-5D)`
`rArrP_(2)=(+5xx0.1)/0.8=5/8=0.625D`
So, option (a) is close to 0.625 as 0.625 `~~` 1 D.