Two similar thin equi-convex lenses, of focal length \(f\) each, are kept coaxially in contact with each other such that the focal length of the combination is \(F_1\)${\mathrm{}}_{}$. When the space between the two lenses is filled with glycerin which has the same refractive index as that of glass \((\mu = 1.5),\) then the equivalent focal length is \(F_2\)${\mathrm{}}_{}$. The ratio \(F_1:F_2\) will be:
1. \(3:4\)
2. \(2:1\)
3. \(1:2\)
4. \(2:3\)

A double convex lens has a focal length of \(25\) cm. The radius of curvature of one of the surfaces is double of the other. What would be the radii if the refractive index of the material of the lens is \(1.5?\)
1. \(100\) cm, \(50\) cm
2. \(25\) cm, \(50\) cm
3. \(18.75\) cm, \(37.5\) cm
4. \(50\) cm, \(100\) cm

Two identical thin plano-convex glass lenses (refractive index = \(1.5\)) each having radius of curvature of \(20\) cm are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of a refractive index of \(1.7\). The focal length of the combination is:
1. \(-20\) cm
2. \(-25\) cm
3. \(-50\) cm
4. \(50\) cm

When a biconvex lens of glass having a refractive index of \(1.47\) is dipped in a liquid, it acts as a plane sheet of glass. The liquid must have a refractive index:

A plane convex lens \((\mu= 1.5)\) has a radius of curvature \(10~\text{cm}\). It is silvered on its plane surface. The focal length of the lens after silvering is: