The focal length of an equi-convex lens is greater than the radius of curvature of any of the surfaces. Then, the refractive index of the material of the lens is

Options

A. Greater than zero but less than 1.5
B. Greater than 1.5 but less than 2.0
C. Greater than 2.0 but less than 2.5
D. Greater than 2.5 but less than 2.0

Answer

A

Explanation

From Lens Makers formula, `1/f =-(mu-1)(1/R_(1) – 1/R_(2))`
For equi-convex lens,` R_(1)= + R, R_(2)=- R`
`implies 1/f= (mu-1)(2/R)`
`impliesf= R/(2(mu-1))`
If `f lt R` then, `2(mu-1) lt 1`
`implies (mu-1) lt 1/2`
`implies(mu-1) lt 0.5`
`implies mu lt 1.5`
Focal length of a convex lens is positive.