Question

. Find the height of an image for an object that is 5.00 in. tall. The object is located 10.0 in. from a convex mirror
and its image is located -5.00 in. from the rear of the mirror

Answer 1

Answer:

"toofani industries "

Answer 2

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given :- }}}$

$\implies\textsf{ Height of the image is 5 inch.} \\\implies\textsf{ Object is located at 10cm inch from mirror .}\\\implies\textsf{ Image is formed 5 inch from rear of Mirror. }$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Find :- }}}$

$\implies\textsf{ The height of the image .}$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Answer :- }}}$

Here since the image is formed in rear of the mirror , hence According to sign convention the image distance will be positive , object distance will be negative and height of the image / object will be Positive ( since convex mirror forms only virtual and erect images of real objects. ) We know that magnification is defined as height of the image by the height of the object or image distance by object distance.

$\underline{\purple{\boldsymbol{ Using \ formula \ of \ magnification :- }}}$

$\sf:\implies \pink{ Magnification = \dfrac{height_{image }} {height_{object}}=\dfrac{ Distance_{image}}{Distance_{object}} }\\\\\sf:\implies \dfrac{ h_i}{h_o} =\dfrac{v}{u}\\\\\sf:\implies \dfrac{h_i}{5 \ in. } = \dfrac{- 5 \ in. }{-10\ in.} \\\\\sf:\implies h_i = \dfrac{1}{2}\times 5 in. \\\\\sf:\implies\boxed{\pink{\mathfrak{ Height_{(image)} = 2.5 \ inch }}}$

$\underline{\blue{\sf Hence \ the \ Image \ height\ is \ \textsf{\textbf{-2.5 \ inch }}. }}$