$A$ 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:

$A$ converging lens is used to form an image on a screen. When the upper half of the lens is covered by an opaque screen,

An equiconvex lens made of glass of refractive index $\frac{3}{2}$ has focal length $f$ in air. It is completely immersed in water of refractive index $\frac{4}{3}$. The percentage change in the focal length is

$A$ lens having refractive index $1.6$ has a focal length of $12 \ cm$ when it is in air. Find the focal length of the lens when it is placed in water. (Take the refractive index of water as $1.28$) (in $mm$)

$A$ convex lens makes a real image $4 \, cm$ long on a screen. When the lens is shifted to a new position without disturbing the object,we again get a real image on the screen which is $16 \, cm$ tall. The length of the object must be.......$cm$

$A$ plano-convex lens having a radius of curvature of the first surface $2 \ cm$ exhibits a focal length of $f_1$ in air. Another plano-convex lens with a first surface radius of curvature $3 \ cm$ has a focal length of $f_2$ when it is immersed in a liquid of refractive index $1.2$. If both lenses are made of the same glass of refractive index $1.5$,the ratio of $f_1$ and $f_2$ will be: