$A$ person uses spectacles of power $+2D$. He is suffering from

The lens used to correct long-sightedness (hypermetropia) is:

$A$ short-sighted person can see distinctly only those objects which lie between $10\, cm$ and $100\, cm$ from him. The power of the spectacle lens required to see a distant object is

$A$ myopic adult has a far point at $0.1\, m$. His power of accommodation is $4\, D$.
$(i)$ What power lenses are required to see distant objects?
$(ii)$ What is his near point without glasses?
$(iii)$ What is his near point with glasses? (Take the image distance from the lens of the eye to the retina to be $2\, cm$.)

$A$ man can see objects clearly between $15 \ cm$ and $30 \ cm$. He uses a lens to see far objects. Due to the lens used,the new near point will be at........$cm$.

For a normal eye,the cornea of the eye provides a converging power of $40\, D$ and the least converging power of the eye lens behind the cornea is $20\, D$. Using this information,the distance between the retina and the cornea-eye lens can be estimated to be ......... $cm$.