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 ssize.​

Answer 1

Answer:

Given:

Height of object, $\sf h_o$ = 5 cm

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:

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

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

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

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

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

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

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

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

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

★ Nature of image:

Image is virtual and erect.

Image formed behind the mirror.

⠀━━━━━━━━━━━━━━━━━━━━━━━━━

Size of image,

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

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

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

$:\implies\sf h_i = - 3 \times 5$

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

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

Answer 2

$\begin{lgathered}{\boxed{{\pink{Given}}}}\;\bigstar\\ \\\end{lgathered}$

Height of object, $\sf h_o = 5cm$

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 :-

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

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

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

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

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

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

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

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

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

★ Nature of image:

• Image is virtual and erect.

• Image formed behind the mirror.

⠀━━━━━━━━━━━━━━━━━━━━━━━━━

Size of image,

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

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

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

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

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

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