A thin convergent glass lens `(mu_g=1.5)` has a power of `+5.0D.` When this lens is immersed in a liquid of refractive index `mu_1,` it acts as a divergent lens of focal length `100 cm.` The value of `mu_1` is

Options

A. `4//3`
B. `5//3`
C. `5//4`
D. `9//5`

Answer

B

Explanation

Power =5 `rArrf=20cm`
`1/f=((mu_(g)-mu_(a))/(mu_(a)))(1/(R_(1))-1/(R_(2)))rArr1/20=((1.5)/1-1)(1/(R_(1))-1/(R_(2)))` …(i)
and `1/(-100)=((1.5-mu_(l))/(mu_(l)))(1/(R_(1))-1/(R_(2)))` …(ii)
Dividing I by II we get `-5=(1/2)/(((1.5)/(mu_(l))-1))rArr1-(1.5)/(mu_(l))=1/10rArr9/10=3/(2mu_(l))`
`rArrmu_(l)=5/3`