When a biconvex lens of glass having refractive index $1.47$ is dipped in a liquid, it acts as a plane sheet of glass. This implies that the liquid must have a refractive index

$A$ thin converging lens of focal length $f=25 \,cm$ forms the image of an object on a screen placed at a distance of $75 \,cm$ from the lens. The screen is moved closer to the lens by a distance of $25 \,cm$. The distance through which the object has to be shifted, so that its image on the screen is sharp again, is (in $\,cm$)

$A$ magnifying lens has a focal length of $10 \, cm$. If an image is to be formed at a distance of $30 \, cm$ from the lens,the object should be placed at a distance of ..... $cm$.

$A$ concave lens of glass,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

$A$ biconvex lens $(R_{1}=R_{2}=20 \ cm)$ has a focal length equal to the focal length of a concave mirror. The radius of curvature of the concave mirror is (Refractive Index of glass lens $= 1.5$) (in $cm$)

$A$ 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: