$A$ converging beam of rays is incident on a diverging lens. Having passed through the lens,the rays intersect at a point $15\, cm$ from the lens on the opposite side. If the lens is removed,the point where the rays meet will move $5\, cm$ closer to the lens. The focal length of the lens is $......\, cm$.

The focal length of a thin converging lens in air is $20 \ cm$. When the lens is immersed in a liquid,it behaves like a concave lens of power $1 \ D$. If the refractive index of the material of the lens is $1.5$,the refractive index of the liquid is

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

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:

An object is placed at a distance of $1.50 \, m$ from a screen. $A$ convex lens is placed between them to form a real image four times larger than the object on the screen. The position of the lens is at a distance of ..... (in $, m$)

$A$ glass convex lens $(\mu_g = 1.5)$ has a focal length of $8\, cm$ in air. What will be its focal length when it is immersed in water (in $, cm$)? $(\mu_w = 1.33)$