The distance between an object and its two times magnified real image produced by a convex lens is $45 \,cm$. The focal length of the lens used is . . . . . . $cm$.

$A$ convex lens of radii of curvature $6 \ cm$ and $12 \ cm$ is immersed in a liquid of refractive index $1.3$. If the refractive index of the material of the lens is $1.5$,then the focal length of the lens when immersed in the liquid is (in $cm$)

$(i)$ 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 $= 1.5$)

$A$ concave lens of glass,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:

The refractive index of a converging lens is $1.4$. What will be the focal length of this lens if it is placed in a medium of the same refractive index? (Assume the radii of curvature of the faces of the lens are $R_{1}$ and $R_{2}$ respectively)

Assertion: Goggles have zero power.
Reason: Radius of curvature of both sides of the lens is same.