Question

an object 2 cm in size is placed 30 cm in front of a concave mirror of focal length 15 CM .at what distance from the mirror should a screen he placed in order to obtain a sharp image ?what will be the nature and the size of the image formed?draw a ray diagram to show the formation of the image in this case?

Answer 1

u= -30

h= 2

F= -15

1/v+1/u=1/F

1/v=1/F-1/u

1/v=1/-15 -1/-30

1/v=1/-15+1/30

1/v=-15+30/-450

1/v=15/-450

v=-450/15

v=-30cm

m=h'/h=-v/u

h'/2=-(-30)/-30

h'/2=30/-30

-30h'=60

h'=60/-30

h'=-2cm

The image is real and inverted.The size of image is equal to the object size

Answer 2

HELLO DEAR,



GIVEN:-
size of object h = +2cm
distance of object u = -30cm
focal length f = -15cm


by mirror formula, 1/v + 1/u = 1/f

=> 1/v = 1/f - 1/u

=> 1/v = 1/(-15) - 1/(-30)

=> 1/v = 1/30 - 1/15

=> 1/v = (1 - 2)/30

=> 1/v = -1/30

=> v = -30cm


the screen should be placed at 30cm in front of the mirror so, we obtain a real image


magnification, m = h'/h = -v/u
where, h' = size of image


then, h'/h = -v/u

=> h'/2 = -(-30/-30)

=> h' = -2cm

hence, the image formed is real and the same size as object



I HOPE IT'S HELP YOU DEAR,
THANKS