Question

an object 5cm in length is placed at a distance of 20cm in front of a convex mirror of radius of curvature 30cm . find the position of the image,its nature and size!??​

Answer 1

Answer:

virtual and erect

enlarged

big

Answer 2

↬Given :-

• Object distance (u) = -20 cm

• Radius of curvature (R) = 30 cm

• Height of object (h1) = 5 cm

⸕ we know that,

➝ F = R/2

➝ F = 30/2

➝ F = 15 cm

↬ To find :-

• The position, nature and size of the image

↬ Solution :-

$\pink{\bigstar}\:\:\underbrace{\bf\purple{{By\:using\:mirror\:formula}}}$

$\green{\bigstar}\:\:{\underline{\boxed{\bf\red{\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}}}}}$

$\rm\longrightarrow\:\dfrac{1}{v}+\dfrac{1}{-20}=\dfrac{1}{15}$

$\rm\longrightarrow\:\dfrac{1}{v}-\dfrac{1}{20}=\dfrac{1}{15}$

$\rm\longrightarrow\:\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{20}$

$\rm\longrightarrow\:\dfrac{1}{v}=\dfrac{4+3}{60}$

$\rm\longrightarrow\:\dfrac{1}{v}=\dfrac{7}{60}$

$\rm\longrightarrow\:v=\dfrac{60}{7}$

$\rm\longrightarrow\:v=8.75\:cm$

Hence,the position of image will be 8.75 cm behind the mirror on its right side.

and nature of the convex mirror will be virtual and erect.

Now,

$\blue{\bigstar}\:\:{\underline{\boxed{\bf\orange{Magnification\:(m)=-\dfrac{v}{u}}}}}$

$\rm\longrightarrow\:m=-\dfrac{8.57}{-20}$

$\rm\longrightarrow\:m=0.42$

Then,

$\orange{\bigstar}\:\:{\underline{\boxed{\bf\pink{m=\dfrac{h_2}{h_1}}}}}$

$\rm\longrightarrow\:0.42=\dfrac{h_2}{5}$

$\rm\longrightarrow\:h_2=0.42\times\:5$

$\rm\longrightarrow\:h_2=2.1\:cm$

Hence,the size of image will be 2.1 cm.