Question

An object 5 cm in length is placed at a distance of 20cm in front of concave mirror of radius of curvature 30cm.Find the position of the image‚ its nature and size.​

Answer 1

$\begin{gathered}\dag\;{\underline{\frak{Using\; Formula\;of\;magnification\;:}}}\\ \\\end{gathered}$

$\begin{gathered}\star\;{\boxed{\sf{\purple{ \dfrac{h_i}{h_o} = \dfrac{v}{u}}}}}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf \dfrac{h_i}{5} = \cancel{ \dfrac{- 60}{20}}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf h_i = - 3 \times 5\\ \\\end{gathered}$

$\begin{gathered}:\implies{\boxed{\frak{\pink{h_i = - 15\;cm}}}}\;\bigstar\\ \\\end{gathered}$

$\therefore\;{\underline{\sf{Height\;or\;size\;of\;image\;is\; \bf{- 15\;cm}.}}}$

Answer 2

GIVEN :-

• Height of the object = 5cm.
• Object Distance (u) = -20cm.
• Radius of curvature (r) = -30cm.

So, f = -30/2 = -15

TO FIND :-

• Position, Size and Nature of the image.

Formula to be used :-

$\huge \fbox \blue{1/v-1/u=1/f}$

Solution:-

➣ 1/v = 1/f+1/u

➣ 1/v = 1/-15+1/-20

➣ 1/v = -4+3 /60

➣ 1/v = -1/60

➣ v = -60 cm.

• The image is at a dist. of 60cm. in front of the mirror.
• This shows that the image is beyond C.
• It's -ve sign shows that it is real and inverted.

➠Size of the image:-

Size of the image:-Formula to be used here:-

$\huge \fbox \blue{m= -v/u}$

Where,

• M = magnification
• V = image Distance
• U = object distance
• \implies M = -(-60)/-30

\implies m = -2

➢When, m is greater than 1 it shows that the image is greater than the object.

➢When, m is greater than 1 it shows that the image is greater than the object.Hence, the image is enlarged as well.