The refractive index of a glass convex lens is $1.5$. The radius of curvature of each of the two surfaces of the lens is $40 \ cm$. The ratio of the power of the lens when immersed in a liquid of refractive index $1.25$ to that when placed in air is:

The radius of curvature for a convex lens is $40 \ cm$ for each surface. Its refractive index is $1.5$. The focal length will be ...... $cm$.

Where should an object be placed on the axis of a convex lens of focal length $8 \ cm$ so as to achieve magnification of $-4$ (in $cm$)? (Distances are measured from the optic centre of the lens)

$A$ thin lens made of glass of refractive index $\mu_g = 1.5$ has a focal length of $12 \ cm$ in air. If it is immersed in water of refractive index $\mu_w = 4/3$,what will be its new focal length in $cm$?

$A$ convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its focal length will

$(i)$ If $f=0.5 \,m$ for a glass lens,what is the power of the lens?
$(ii)$ The radii of curvature of the faces of a double convex lens are $10 \,cm$ and $15 \,cm$. Its focal length is $12 \,cm$. What is the refractive index of glass?
$(iii)$ $A$ convex lens has $20 \,cm$ focal length in air. What is its focal length in water? (Refractive index of water $= 1.33$,refractive index of glass $= 1.5$)