An object 4 cm high is placed what 40.0 CM in front of the concave mirror of focal length 20 cm find the distance from the mirror at which screen is placed in order to obtain a sharp image.Also,find the nature of the image formed.
Answers
Answer:
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Answer:
The distance of the image at 40 cm from the mirror and the image is inverted.
Explanation:
Explanation:
Given that,
Height of the object h = 4 cm
Distance of the object u = -40 cm
Focal length f = - 20 cm
Using mirror's formula
\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}
f
1
=
v
1
+
u
1
\dfrac{1}{-20}=\dfrac{1}{v}+\dfrac{1}{-40}
−20
1
=
v
1
+
−40
1
\dfrac{1}{v}=-\dfrac{1}{40}
v
1
=−
40
1
v = -40\ cmv=−40 cm
The distance of the image is 40 cm.
The magnification is
m = -\dfrac{-v}{-u}m=−
−u
−v
m =\dfrac{40}{40}m=
40
40
m=-1m=−1
We know that,
m=\dfrac{h'}{h}m=
h
h
′
-1=\dfrac{h'}{h}−1=
h
h
′
h' = -4\ cmh
′
=−4 cm
Hence, The distance of the image at 40 cm from the mirror and the image is inverted.