An object is placed exactly midway betweena concave mirror of radius of curvature 30 cm. The mirrors face each other and are 50 cm part. Determine the nature and position of the image formed by the successive reflections, first at the concave mirror and then at the convex mirror
plz answer my question it's urgent
Answers
answr
search
What would you like to ask?
12th
Physics
Ray Optics and Optical Instruments
Combination of Lenses
An object is placed exactly...
PHYSICS
An object is placed exactly midway between a concave mirror of radius of curvature 40cm and a convex mirror of radius of curvature 30cm. The mirrors face each other and are 50cm apart. Determine the nature and position of the image formed by successive reflections first at the concave mirror and then at the convex mirror.
769507
MEDIUM
Share
Study later
ANSWER
Given: An object is placed exactly midway between a concave mirror of radius of curvature 40cm and a convex mirror of radius of curvature 30cm. The mirrors face each other and are 50cm apart.
To find the nature and position of the image formed by successive reflections first at the concave mirror and then at the convex mirror.
Solution:
For concave mirror,
As per the given condition,
object distance, u=25cm, as the mirrors are 50cm apart.
Radius of curvature, R=40cm
So the focal length, f=
2
R
=
2
40
=20cm
Applying lens formula, we get
f
1
=
v
1
+
u
1
⟹
v
1
=
f
1
−
u
1
⟹
v
1
=
20
1
−
25
1
⟹
v
1
=
100
5−4
⟹v=100cm
Therefore the image is 100cm behind the concave mirror.
Magnification, m
1
=−
u
v
=−
25
100
=−4
Hence the image formed by the concave mirror is inverted, virtual and enlarged.
This becomes object for convex mirror, so
object distance, u
′
=−(100−50)=−50cm
Radius of curvature of convex mirror, r=30cm
So the focal length, f
′
=
2
r
=
2
30
=−15cm
Applying lens formula, we get
f
′
1
=
v
′
1
+
u
′
1
⟹
v
′
1
=
f
′
1
−
u
′
1
⟹
v
′
1
=
−15
1
−
−50
1
⟹
v
′
1
=
150
−10+3
⟹v=−21.4cm
Hence the position of the final image is 21.4 behind the convex mirror.
Magnification, m
2
=−
u
′
v
′
=−
−50
−21.4
=−0.43
So the nature of the final image formed is inverted of inverted, i.e., upright.
MARK AS BRAINLIEST ANSWER...