A convex lens of refractive index 3/2 has a power of `2.5^(@)`. If it is placed in a liqud of refractive index 2,the new power of the lens is

Options

A. -1.25D
B. -1.5D
C. 1.25 D
D. 1.5D

Answer

1

Explanation

p = `1/f`
p = `1/f = 1/2.5 = 0.4m`
`u_(g)=1.5`
`1/f = (mu-1)(1/R_(1)-1/R_(2))`
`1/0.4 = 0.5 (1/R_(1)-1/R_(2))`
`1/0.4 = (1/R_(1)-1/R_(2))`
`1/f. = (mu_(2)/mu_(1)-1)(1/R_(1)-1/R_(2))`
` (1.5/2 - 1)1/0.2`
= -`0.5/2 xx 1/0.2 `=-1.5D