Question

A screen bearing a real image of
magnification mı, formed by a
convex lens, is moved by a distance
x. The object is then moved until a
new image of magnification m, is
formed on screen. The focal length
of lens is :​

Answer 1

focal length of lens f = $\frac{x}{(m_1- m})$

Real images are obtained by shifting a convex lens between an object and screen separated by a fixed distance (D)

Essential condition: D > 4f

where 'f' is the focal length of convex lens.

initial magnification of image = m₁

magnification of image after shift = m

shift distance (S) = x

focal length (f) = $\frac{(Distance)^2-(shift)^2}{4D}$

f = $\frac{D^2-S^2}{4D}$ by solving this equation we get

f = $\frac{s}{m_1-m}$