An equiconvex lens of glass of focal length $0.1 \text{ m}$ is cut along a plane perpendicular to the principal axis into two equal parts. The ratio of the focal lengths of the new lenses formed is

$A$ thin biconvex lens is prepared from glass $(mu = 1.5)$ where both curved surfaces have equal radii of $20 \text{ cm}$ each. The left side surface of the lens is silvered from the outside to make it reflecting. To have the position of the image and the object at the same place,the object should be placed from the lens at a distance of . . . . . . $\text{cm}$.

An object is placed $0.1 \,m$ in front of a convex lens of focal length $20 \,cm$ made of a material of refractive index $1.5$. The surface of the lens away from the object is silvered. If the radius of curvature of the silvered surface is $22 \,cm$,then the distance of the final image from the silvered surface is (in $\,cm$)

In the diagram given below,there are three lenses formed. Considering the negligible thickness of each of them as compared to $|R_1|$ and $|R_2|$,i.e.,the radii of curvature for the upper and lower surfaces of the glass lens,the power of the combination is:

$A$ point object is held above a thin equiconvex lens at its focus. The focal length is $0.1 \ m$ and the lens rests on a horizontal thin plane mirror. The final image will be formed at

The figure shows an equiconvex lens (of refractive index $1.50$) in contact with a liquid layer on top of a plane mirror. $A$ small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be $45.0 \; cm$. The liquid is removed and the experiment is repeated. The new distance is measured to be $30.0 \; cm$. What is the refractive index of the liquid?