Question

12. An object of height 2cm is placed in front of a concave mirror at 12cms distance. Find the
image distance and the height of the image formed if its Radius of curvature is 24cms.​

Answer 1

Answer:

12cm and 2 cm

Explanation:

Focal length= 24cm/2 = 12cm

Because, the object is placed at the focal lenght.

Therefore,

Image will be of same size(2cm), Inverted, Real and formed at focus.

Answer 2

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

$\textsf{ \to An object of height 2cm is placed in front of a concave mirror.}\\\textsf{ \to Object is kept at a distance of 12 cm.}\\\textsf{ \to Radius of curvature of lens is 24 cm. }$

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

$\textsf{ \to The image distance and the height of the image.}$

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

Given that the Radius of Curvature is 24 cm . Hence we know that the radius of curvature is double of the focal length of the mirror . Hence here the focal length will be 24cm÷2 = 12cm. Now the Object distance is 12cm . This means the object is placed at focus . Let's use Mirror Formula to find the image distance.

$\underline{\purple{\sf\mathscr{U}sing \ \mathscr{M}irror \ \mathscr{F}ormula :- }}$

$\sf:\implies \pink{ \dfrac{1}{v}+\dfrac{1}{u}= \dfrac{1}{f}}\qquad \bigg\lgroup \red{\bf Signs \ have \ usual \ meaning }\bigg\rgroup \\\\\sf:\implies \dfrac{1}{v}+\dfrac{1}{-12cm} = \dfrac{1}{-12cm}\\\\\sf:\implies \dfrac{1}{v}=\dfrac{1}{12cm}+\dfrac{1}{12cm} \\\\\sf:\implies \dfrac{1}{v} = 0 \\\\\sf:\implies v =\dfrac{1}{0}\\\\\sf:\implies \boxed{\pink{\mathfrak{ Image \ Distance \ = \ \infty }}}$

$\underline\textsf{ \blue{ Hence the image is formed at \textbf{Infinity} .}}$

Hence here we can see that the image is formed at infinity. This can also be Proved with the help of the ray diagrams , refer to attachment :) .