Question

an object of 5 cm is placed in front of convex mirror having radius of 30 cm find the position of image and magnification of image??

Answer 1

Correct Question :

An object 5 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image & magnification of image.

Solution :

We have;

• Size of object, (h1) = 5 cm
• Distance of object from mirror, (u) = -20 cm
• Radius of curvature, (R) = 30 cm

A/q

$\longrightarrow\sf{focal\:length=\dfrac{Radius\:of\:curvature}{2}} \\\\\longrightarrow\sf{focal\:length=\cancel{\dfrac{30}{2} }}\\\\\longrightarrow\sf{focal\:length=15\:cm}$

Using formula of the mirror :

$\boxed{\bf{\frac{1}{f} =\frac{1}{v} +\frac{1}{u}}}}$

$\longrightarrow\sf{\dfrac{1}{15} =\dfrac{1}{v}+\dfrac{1}{-20} }\\\\\\\longrightarrow\sf{\dfrac{1}{15} =\dfrac{1}{v}-\dfrac{1}{20} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{15} +\dfrac{1}{20} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{4+3}{60} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{7}{60} }\\\\\\\longrightarrow\sf{7v=60\:\:\underbrace{\sf{cross-multiplication}}}\\\\\longrightarrow\sf{v=\cancel{60/7}}\\\\\longrightarrow\bf{v=8.57\:cm}$

∴ The image is formed behind the mirror at a distance of 8.57 cm .

We know that;

$\mapsto\sf{m=\dfrac{h_2}{h_1} =\dfrac{-v}{u} }\\\\\\\mapsto\sf{\dfrac{h_2}{5} =\dfrac{8.57}{20} }\\\\\\\mapsto\sf{20h_2=8.57\times 5}\\\\\\\mapsto\sf{h_2=\cancel{\dfrac{42.85}{20} }}\\\\\\\mapsto\bf{h_2=2.14\:cm}$

∴ Size of image will 2.14 cm, Image is virtual & erect .

Answer 2

Answer:

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Explanation:

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