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

Step-by-step solution

Focal length of equiconcave lens f = - R 2 μ - 1 Where μ = μ lens μ medium Therefore focal length in the medium f med = - R 2 1 .5 1 .75 - 1 f med = + R 2 0 .25 1 .75 f med = + 3 .5 R Positive sign implies convergent sign.