an object 4cm in size, is placed at 25cm in front of a concavr mirror

1.	an object 4cm in size, is placed at 25cm in front of a concavr mirror of focal length 15cm.at wt distance from the mirror would a screen be placed in order to obtain a sharp image? find the nature and the size of the image
Answer» Given : In concave mirror, Object height = - 4cm. Object DISTANCE = 25cm. Focal length = - 15cm. Remember ! In concave mirror focal length is negative. To find : (a) Image distance (b) Size of image (c) Nature of image Solution : (a) using mirror formula .i.e., The relationship between image distance, object distance and focal length of a spherical mirror is known as Mirror formula. The mirror formula can be written as : $\boxed{ \bf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }$ where, v denotes image distance. u denotes object distance. f denotes focal length. substituting all the given values in the formula, $\leadsto{ \sf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }$ $\leadsto{ \sf \dfrac{1}{v} + \dfrac{1}{( - 25)} = \dfrac{1}{ - 15} }$ $\leadsto{ \sf \dfrac{1}{v} - \dfrac{1}{25} = - \dfrac{1}{15} }$ $\leadsto{ \sf \dfrac{1}{v} = - \dfrac{1}{15} + \dfrac{1}{25} }$ $\leadsto{ \sf \dfrac{1}{v} = - \dfrac{ 7 + 3}{75} }$ $\leadsto{ \sf \dfrac{1}{v} = - \dfrac{10}{75} }$ $\leadsto{ \sf v= - \dfrac{7.5}{10} }$ $\leadsto \underline{\boxed{ \sf v = - 7.5 \: cm}}$ thus, the Image distance is -7.5cm (b) MAGNIFICATION : The magnificent PRODUCED by a mirror is equal to the ratio of the image distance to the object distance with a minus SIGN, $\leadsto{ \bf m= - \dfrac{v}{u} }$ $\leadsto{ \sf m= - \dfrac{( - 7.5)}{25} }$ $\leadsto{ \sf m= 0.3 }$ we know that, $\leadsto{ \bf m= \dfrac{h'}{h} }$ $\leadsto{ \sf 0.3= \dfrac{h'}{4} }$ $\leadsto{ \sf h' = 0.3 \times 4 }$ $\leadsto \underline{\boxed{ \sf h' = 1.2 \: cm }}$ thus, Image height h' is 1.2cm. (c) Nature of image : The image is real and inverted. The image is diminished.