An object 4cm high is placed at a distance
of 6cm in front of a concave mirror
of focal length 12cm. find the position
nature and size of the image
Answers
ho= 4cm
u= -6 cm
f = -12 cm
mirror formula :-
1/f = 1/u +1/v
1/v = 1/f - 1/u
1/v = 1/-12 + 1/6
= -1+2/12 ( take the lcm of 12 and 6)
= 1/12
v = 12 (the value is in positive so, the image is formed on the other side of the mirror)
m = -v / u
= - 12 / - 6
= 2
(the value image formed is virtual and erect)
I HOPE IT'S HELPFUL......
Answer:
The object is kept between the focus and pole.
So,the image is enlarged, virtual and erect.
u=(-6)cm
f=(-12)
From mirror formula,
(-1/6)+(1/v)=(-1/12)
1/v=(-1/12)+1/6
1/v=1/12
v=12cm.
Image is formed 12cm behind mirror.
Magnification=(-12/-6)=2
Size of image=4(2)=8cm.
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