Question

solve plzzzzzzzzzzz..... ​

Answer 1

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

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.

hope this helps you