Question

$\huge\tt{\underline{\pink {MATH \: Chapter \: - \: 8 }}}$


$\huge\tt{\underline{\red {NOTE}}}$

$if \: you \: know \: ASSAMMESE \: Language \: then \: answer \: me \: in \: assammese\: or \: if \: you \: don't \: know \: the \: language\: then\: ok\: you \: can \: give \: answer \: in \: ENGLISH.$


$\huge{\pink{★}$[/tex]\huge{\red{DON'T \: SPAM }}} [/tex]​

Answer 1

Answer:

Answer

Solution 1. Given : object distance (u)= -20 cm

Object size (h)= 5 cm

Radius of curvature = 30

Therefore focal length (f)= 30/2

= 15 cm

object image (h' )= ?

image distance( v)= ?

Now,

1 1 + 1

Position of image (v)= f = u v

1 = 1 -1

/ /= v u f

1 = 1 - 1

/ /= v ( -20 ) 15

/ /= 1 = - 15 (- 20 )

v -300

/= 1 = -35

v -300

/ /= 1 = 7

v 60

60

/ /= v = 7

/ /= v = 8.57 cm

Therefore ,IMAGE is formed behind the mirror.

ALSO,

MAGNIFICATION: m = h' = -v

h u

/ /= h' = -v h

u

/ /=h' = ( -8.75) x 5

-20

/ /=h' = -42.85

-20

/ /=h' = 2.14cm

Therefore NATURE OF IMAGE = VIRTUAL & ERECT

SIZE OF IMAGE = DIMINiSHED .

Explanation:

Attachment tu SWA correct Ase no nai Answer.

Answer 2

It's your answer. ‏‏‎ ‎

bye