$A$ thin lens is made of a material with a refractive index of $\mu = 1.5$. It is a biconvex lens. When it is immersed in water,it behaves as a/an:

An illuminated object and a screen are placed $90 \, cm$ apart. What is the focal length of the lens required to produce an image on the screen,twice the size of the object?.......$cm$

$A$ bi-concave glass lens having refractive index $1.5$ has both surfaces of same radius of curvature $R$. On immersion in a medium of refractive index $1.75$,it will behave as a

When an object is at a distance $u_1$ and $u_2$ from a lens,a real image and a virtual image are formed,respectively,having the same magnification $m$. The focal length of the lens is:

The following figure shows a beam of light converging at point $P$. When a concave lens of focal length $16 \,cm$ is introduced in the path of the beam at a place shown by the dotted line such that $OP$ becomes the axis of the lens,the beam converges at a distance $x$ from the lens. The value of $x$ will be equal to (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 $cm$)