Question

An object of height 6cm is placed at a distance of 10cm from convex mirror with radius of curvature 30cm find the height of it's image​

Answer 1

S O L U T I O N :

$\bf{\large{\underline{\bf{Given\::}}}}}$

• An object of height (h1) = 6 cm.
• Distance of object from mirror (u) = -10 cm.
• Radius of curvature (R) = 30 cm.

$\bf{\large{\underline{\bf{To\:find\::}}}}}$

The height of It's image.

$\bf{\large{\underline{\bf{Explanation\::}}}}}$

We know that formula of the focal length :

$\longrightarrow\sf{Focal\:length\:(f)=\dfrac{Radius\:of\:curvature}{2} }\\\\\\\longrightarrow\sf{Focal\:length\:(f)=\cancel{\dfrac{30}{2}} }\\\\\\\longrightarrow\sf{Focal\:length\:(f)=15\:cm}$

Now;

By using formula of mirror :

$\longrightarrow\sf{\dfrac{1}{f} =\dfrac{1}{v} +\dfrac{1}{u} }\\\\\\\longrightarrow\sf{\dfrac{1}{15} =\dfrac{1}{v} +\bigg(\dfrac{1}{-10} \bigg)}\\\\\\\longrightarrow\sf{\dfrac{1}{15}=\dfrac{1}{v}-\dfrac{1}{10} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{15} +\dfrac{1}{10} }\\\\\\\longrightarrow\sf{\dfrac{1}{v}=\dfrac{2+3}{30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v}=\cancel{\dfrac{5}{30}}}\\\\\\\longrightarrow\bf{v=6\:cm}$

Now;

$\mapsto\bf{m=\dfrac{Height\:of\:image\:(I)}{Height\:of\:object\:(O)} =\dfrac{Distance\:of\:image\:(v)}{Distance\:of\:object\:(u)}}$

$\longrightarrow\sf{m=\dfrac{h_2}{h_1} =\dfrac{-v}{u} }\\\\\\\longrightarrow\sf{\dfrac{h_2}{6} =\dfrac{6}{10} }\\\\\\\longrightarrow\sf{10h_2=6\times6}\\\\\\\longrightarrow\sf{h_2=\cancel{\dfrac{36}{10} }}\\\\\\\longrightarrow\bf{h_2=3.6\:cm}$

Thus;

The height of it's image is 3.6 cm , image is virtual, erect.