u ; Dividing both sides with v, u After rearranging, u ( . ) - - - - Unit ray optics Solution longitudinal magnification m l lengthof image lengthof object ( ) = ( l ´) ( l ) Given: length of object, l = f For the given condition, the image formation is shown in the figure. A' B' u A = v A = 2f = R f/ F u B v B l ' P Rod Image l ' C Image Let, l' be the length of the image, then m f = ′ = ′ ′ = / or) Image of one end coincides with the respective end of object. Thus, the coinciding end must be at centre of curvature.
u u u m f + + ’ Mirror equation, u − = − After simplifying, ;( ; m l = . EXAMPLE . An object is placed at a distance of . cm from a concave mirror of focal length .
cm. (a) What distance from the mirror a screen should be placed to get a sharp image? (b) What is the nature of the image? Solution Given, f = – cm, u = – cm (a) Mirror equation, u Rewriting to find v, u Substituting for f and u, v = − −− v = − −− = − = − cm =− .
The screen is to be placed at distance . cm to the left of the concave mirror. (b) Magnification, m h h u ′ =− h h ′ =−− =− As the sign of magnification is negative, the image is inverted. As the magnitude of magnification is , the image is enlarged three times.
As the image is formed to the left of the concave mirror, the image is real. EXAMPLE . A thin rod of length f / is placed along the optical axis