A thin lens of refractive index 1.5 has optical power of -5 D in air. Its optical power in a liquid medium with refractive index 1.6 is D.

Step-by-step solution

Given, refractive index of lens, _g=1.5 Refractive index of medium, _m=1.6 Power of lens, P=-5 D Focal length of lens, f_l= 1 -5 ~m =-20 ~cm From lens maker formula, we have array r P= 1 f_l = ( _s _a -1 ) ( 1 R₁ - 1 R₂ ) (i) ( _a= . refractive index of air) array P^ = 1 f_m = ( _g _m -1 ) ( 1 R₁ - 1 R₂ )...(ii) Dividing Eq. (ii) by Eq. (i), we get aligned P^ P & = ( _g _m -1 ) ( _g _a -1 ) = 1.5 1.6 -1 1.5 1 -1 & P^ P = -1 16 2 1 P^ = -1 8 (-5) & P^ =0.625 D aligned This is close to option (a).