Suppose an object has thickness du so that it extends from object distance u to `u+du.` Prove that the thickness dv of its image is given by `(-v^2/u^2)du,` so the longitudinal magnification `(dv)/(du)=-m^2,` where m is the lateral magnification.

If f is focal length and u is object distance for spherical mirror, then lateral magnification m = ......

A lens forms real and virtual images of an object, when the object is at u_(1) and u_(2) distance respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as m)

If u=e ^(log cos 4x ),v =e ^(log sin 4x ),then (dv)/(du)=

The object distance u, the image distance v and the magnification m in a lens follow certain linear relations.

if int_(log2)^(x) (du)/(e^u-1)^(1/2) =pi/6 then e^x=