1.

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 isA. 2.5 DB. `-2.5 D`C. `1.25 D`D. `- 1.25 D`

Answer» Correct Answer - D
We have, `P=(mu-1)[(1)/(R_(1))-(1)/(R_(2))]`
`2.5=((3)/(2)-1)((1)/(R_(1))-(1)/(R_(2)))`
`P=((3//2)/(2)-1)((1)/(R_(1))-(1)/(R_(2)))`
`therefore " " P=-1.25 D`


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