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

Answer:

• Radius of curvature (R) = 30 cm
• f = R/2 = 30/2 = 15 cm
• u = -20 cm,
• h= 5 cm.

1/v +1/u = 1/f

1/v = 1/15+ 1/20 = 7/60

v = 60/7 = 8.6 cm.

Image is virtual and erect and formed behind the mirror.

hi/h0= v/u

hi/5= 8.6/20

hi = 2.2 cm.

Size of image is 2.2 cm.

Answer 2

Given:

• Object distance, u = - 20 cm
• Radius of curvature, R = 30 cm
• Focal length, f = R/2 = 30/2 = - 15 cm

To Find :

Position of image, it's nature and size?

Solution:

$\begin{gathered}\dag\;{\underline{\frak{Using\;mirror\;formula\;:}}}\\ \\\end{gathered}$

$\begin{gathered}\star\;{\boxed{\sf{\purple{ \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}}}}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf \dfrac{1}{- 15} = \dfrac{1}{v} + \dfrac{1}{- 20}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf \dfrac{1}{v} = \dfrac{1}{- 15} - \dfrac{1}{ - 20}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf \dfrac{1}{v} = \dfrac{1}{- 15} + \dfrac{1}{20}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf \dfrac{1}{v} = \dfrac{- 4 + 3}{60}\\ \\\end{gathered}$

$\begin{gathered}:\implies\sf \dfrac{1}{v} = \dfrac{- 1}{60}\\ \\\end{gathered}$

$\begin{gathered}:\implies{\boxed{\frak{\pink{v = - 60\;cm}}}}\;\bigstar\\ \\\end{gathered}$

$\begin{gathered}\therefore\;{\underline{\sf{Image\;distance\;is\; \bf{- 60\;cm}.}}}\\ \\\end{gathered}$

★ Nature of image:

Image is virtual and erect.

Image formed behind the mirror.

⠀━━━━━━━━━━━━━━━━━━━━━━━━━

Size of image,

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

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

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

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

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

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