An object, 4.0 cm in size, is placed at 25.0 cm in front of a mirror of focal length 15.0 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and the size of the image
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Explanation:
Given:
The size of object = 4.0cm
Distance of the object = 25cm (-ve as it is in front of mirrors)
focal length of concave mirror = 15cm (-ve)
Distance of the image is given by-
v
1
+
u
1
=
f
1
v
1
+
−25
1
=
−15
1
v
1
=
−15
1
+
25
1
=
150
−10+6
150
−4
⇒v=−37.5cm
The screen should be at 37.5cm in front of the lens.
(ii)
The size of the image is given by-
h
0
h
i
=
u
−v
⇒h
i
=
u
−v×h
0
=
−25
−(−37.5)×4
h
i
=−
25
150
=−6cm
The image will be 6cm high and it will be inverted.
(iii)
Refer image 1
The image will be formed at a distance of 37.5cm from the mirror.
It will be an inverted image
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