A concave lens of glass, refractive index 1.5 has both surfaces of same radius of curvature R. On immersion in a medium of refractive index 1.75, it will behave as a

Options

A. convergent lens of focal length `3.5 R`
B. convergent lens of focal length `3.0 R `
C. divergent lens of focal length 3.5 R
D. divergent lens of focal length 3.0 R

Answer

A

Explanation

`R_(1) = -R , R_(2) = +R, mu_(g) = 1.5 "and" mu_(m) = 1.75`
`therefore (1)/(f) = ((mu_(g))/(mu_(m)) - 1) ((1)/(R_(1)) - (1)/(R_(2)))`

Subsituting the values,
we have `(1)/(f) = ((1.5)/(1.75) - 1) ((1)/(-R) - (1)/(R)) = (1)/(3.5 R)`
`therefore f = +3.5 R`
Therefore , in the medium it will behave like a convergent lens of focal length `3.5 R` . it can be understood as , `mu_(m) gt mu_(g)` , the lens will change its behaviour