Double convex lenses are to be manufactured from a glass of refractive index $1.55$ with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be $20 \ cm$ (in $cm$)?

$A$ convex lens of focal length $25 \ cm$ and made of glass with refractive index $1.5$ is immersed in water. The absolute change in focal length of the glass is [Use refractive index of water = $\frac{4}{3}$] (in $cm$)

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 the material of the lens?

Given below are two statements:
Statement $(I)$: When an object is placed at the centre of curvature of a concave lens,the image is formed at the centre of curvature of the lens on the other side.
Statement $(II)$: $A$ concave lens always forms a virtual and erect image.
In the light of the above statements,choose the correct answer from the options given below:

The radii of curvature of both the surfaces of a convex lens of focal length $f$ and focal power $P$ are equal. One of the surfaces is made plane by grinding. The new focal length and focal power of the lens are:

$A$ bi-convex lens made of glass (refractive index $1.5$) is put in a liquid of refractive index $1.7$. Its focal length will