Question

an object of 4.0 cm in size is placed at 25 cm in front of concave mirror of a radius of curvature is 30 cm at what distance from the mirror should a screen be placed in order to obtain a sharp image . find the nature of size of image.?

Answer 1

Concave mirror.
height = h  = 4 cm
distance of object from the mirrror  =  25 cm =>  u = - 25 cm
R = 30 cm  => focal length = R/2 =>     f =  - 15 cm

   1/v  + 1/u  =  1/f    =>    1/v  =  1/(-15)  -  1/(-25)  =  - 1/15 + 1/25
   =>  1/v = - 2 / 75        =>  v = - 37.5 cm
The object forms an image at 37.5 cm from the center of mirror in the direction opposite to incident rays.  ie., image is formed behind the object.    |v| > R and |u|>f , So real image is formed.
Magnification =    v / u  =  37.5 / -25  = - 1.5
So image is real and inverted.

Answer 2

Frome given statement , the Object is hO = + 4.0 cm; The Object is palced at a of distance u D= – 25.0 cm; The concave mirror radius = 30 cm so, the Focal length of the mirror will be F = –15.0 cm; //To find out the screen placement for sharp image,let's represent as M// The formula to find the screen distance or screen placement for sharp image is, 1/M+ 1/D= 1/F. //rearrange the formula// 1/ν = 1/F - 1/D= - ( 1/15cm) - ( - (1/25cm) ) = -1/15.0 + 1/25.0 = - (2 /75 ) = - 37.5 cm To fund out the height of the image, let's represent by H The formula to find height of an image is, H/O = - (M/D). // rearrange the formula// or H = - (MO)/D = - { [ (37.5)(4) ] / 25 } = -6cm