an object of 4 cm height is placed at a distance of 18 cm from concave mirror having focal length 12 cm
Answers
Answer:
AnSwEr
Explanation:
The image is formed at the distance of 24 cm from the mirror and the image is virtual and erect. The height of the image is 8 cm.
Explanation:
Given that,
Height of the object h = 4 cm
Distance of the object u= -12 cm
focal length f = -24 cm
Using mirror's formula
\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}
f
1
=
v
1
+
u
1
Put the value of f and u into the formula
\dfrac{1}{-24}=\dfrac{1}{v}+\dfrac{1}{-12}
−24
1
=
v
1
+
−12
1
\dfrac{1}{f}=\dfrac{1}{24}
f
1
=
24
1
v = 24\ cmv=24 cm
The image is formed behind the mirror.
The formula of magnification
m = -\dfrac{v}{u}=\dfrac{h'}{h}m=−
u
v
=
h
h
′
\dfrac{24}{12}=\dfrac{h'}{4}
12
24
=
4
h
′
h'=8\ cmh
′
=8 cm
The height of the image is 8 cm.
The image is virtual and erect.
Hence, The image is formed at the distance of 24 cm from the mirror and the image is virtual and erect. The height of the image is 8 cm.
Answer:
What is the question mate?
Explanation:
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