$A$ convex lens of refractive index $\frac{3}{2}$ has a power of $2.5 \ D$. If it is placed in a liquid of refractive index $2$, the new power of the lens is: (in $D$)

For a given lens,the magnification was found to be twice as large when the object was $0.15 \ m$ distant from it as when the distance was $0.2 \ m$. The focal length of the lens is.......$m$.

The slit of a collimator is illuminated by a source as shown in the adjoining figures. The distance between the slit $S$ and the collimating lens $L$ is equal to the focal length of the lens. The correct direction of the emergent beam will be as shown in figure:

When a convex lens is immersed in two different liquids of refractive indices $1.25$ and $1.5$,the ratio of the focal lengths of the lens is $5:16$. The refractive index of the material of the lens is

$A$ double convex lens of glass of $\mu = 1.5$ has a radius of curvature of each of its surfaces equal to $0.2 \ m$. The power of the lens is:

$A$ convex lens of refractive index $1.5$ and focal length $f = 18 \ cm$ is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $\alpha \times f$. The value of $\alpha$ is . . . . . . . (Refractive index of water $= 4/3$)