The power (in diopters) of an equiconvex lens with radii of curvature of $10 \ cm$ and refractive index of $1.6$ is

$A$ convex lens is placed between an object and a screen which are at a fixed distance $D$ apart. For one position of the lens,the magnification of the image formed on the screen is $m_1$. When the lens is shifted by a distance $d$,the magnification of the image formed on the same screen is $m_2$. The focal length of the lens is:

$A$ bi-concave symmetric lens made of glass has a refractive index of $1.5$. It has both surfaces with the same radius of curvature $R$. On immersion in a liquid of refractive index $1.75$,it will behave as a:

Assertion: $A$ double convex lens $(\mu = 1.5)$ has a focal length of $10 \, cm$. When the lens is immersed in water $(\mu = 4/3)$,its focal length becomes $40 \, cm$.
Reason: $\frac{1}{f} = \left( \frac{\mu_l - \mu_m}{\mu_m} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

$A$ convex lens has its radii of curvature equal. The focal length of the lens is $f$. If it is divided vertically into two identical plano-convex lenses by cutting it, then the focal length of the plano-convex lens is ($\mu =$ the refractive index of the material of the lens).

An object is placed in front of a convex lens,and an image of height $8 \, cm$ is formed. When the lens is moved to a new position,the image height becomes $2 \, cm$. Find the height of the object in $cm$.